Questão 12

**Deriving roots options** I need to derive some concise options for roots while being rigorous. I'm looking at quartic integer roots with a product of -21, considering odd factors like ±1, ±3, ±7, and ±21. For the roots to be distinct in mod 5, I determined I could only use 1, 1, 3, and 7 based on monic polynomial constraints. The possible sets have either positive or negative products, but I need to show three steps for clarity.

**Determining integer roots for -21** I’m looking at the possibility of integer roots that multiply to -21. Each root needs to divide -21. I should create a list of divisors and their residue classes. The divisors like ±1, ±3, and ±7 could be relevant here. I want to check residues for various classes too. It seems I need to pick combinations carefully to ensure their product equals 21, particularly considering specific middle choices and how they affect the overall product.

With four integer roots of p(x), their product must equal -21 since that's the constant term for a monic quartic. I'm thinking about working modulo 5, since p(n) must vanish mod 5 whenever n isn't divisible by 5 — this means the roots must cover all nonzero residues mod 5 in some pattern. I'm checking divisors of 21 and their residues mod 5: 1→1, -1→4, 3→3, -3→2, 7→2, -7→3, 21→1, -21→4. I need to find a combination of four integers multiplying to -21 whose residues are exactly {1,2,3,4} in some order, and the set {1,-1,3,7} works since it gives residues 1,4,3,2 — I'm now verifying its product equals -21. I found another valid combination: {1,-1,-3,-7} also has distinct residues {1,4,2,3} and multiplies to -21. Everything else with repeated values fails since those repeats collide on the same residue, so now I'm narrowing down to these two candidate root sets and starting to compute a²+b²+c² for the first case with roots 1, -1, 3, 7. Expanding this polynomial gives x⁴ −10x³ +20x² +10x −21, so a²+b²+c² = 100+400+100 = 600. Checking the second case with roots 1, -1, -3, -7 gives the same expansion pattern with coefficients 10, 20, -10, yielding the identical sum of 600 — so the answer is B.

I'm working through a degree-4 monic polynomial with four integer roots whose product is -21, and reasoning that since p(n) is divisible by 5 for all n not a multiple of 5, the roots mod 5 must hit every nonzero residue class, meaning p(x) mod 5 should factor like (x-1)(x-2)(x-3)(x-4). Checking divisors of 21 against their residues mod 5, I find the root set {1,1,3,7} works with the right sign pattern to give the needed residues {1,4,2,3} and product -21. Testing roots 1, -1, 3, 7, I multiply (x²-1)(x²-10x+21) to get the expanded polynomial. Computing a²+b²+c² for both valid sign cases gives 100+400+100=600 each time, confirming the answer is 600.

Since p(x) is monic degree 4 with roots r1-r4, the constant term gives product of roots equal to -21. For p(n) to be divisible by 5 whenever n isn't a multiple of 5, I need each nonzero residue class mod 5 to be covered by exactly one root — so the four roots must hit residues 1, 2, 3, 4 mod 5, one each. Testing factorizations of -21 into four integers, the case with absolute values 1,1,1,21 fails since three ±1 roots can't give distinct residues. The working case uses |values| 1,1,3,7: pairing 1,-1 for residues 1,4, and either (3,7) or (-3,-7) for residues 2,3, both giving valid sign products of -21. Expanding (x²-1)(x²-10x+21) for roots 1,-1,3,7 gives a=-10, b=20, c=10, and the mirror case with -1,-3,-7 gives a=10, b=20, c=-10 — both yield a²+b²+c² = 600.

The problem is in Portuguese. I will read it carefully. The polynomial has four integer roots. The constant term is -21, so the product of the roots is -21. I will use this to find a² + b² + c². The roots are integers whose product is -21. The condition means p(n) ≡ 0 (mod 5) for all n not divisible by 5. Using Fermat's little theorem, x⁴ ≡ 1 (mod 5) for any integer not divisible by 5. Since -21 ≡ -1 ≡ 4 (mod 5), the condition p(n) ≡ 0 (mod 5) simplifies to a n³ + b n² + c n ≡ 0 (mod 5) for all n not divisible by 5. The quadratic a x² + b x + c must vanish at four distinct nonzero elements of F₅. Since a nonzero polynomial of degree at most two has at most two roots, this forces a ≡ b ≡ c ≡ 0 (mod 5). The next step is to verify whether these conditions are sufficient. The condition is necessary and sufficient. Because the roots are integers, a, b, and c are integers that are multiples of 5. The condition is sufficient. I am now checking whether the roots must cover all nonzero residues modulo 5, and how that affects the sum of products. The condition is fully equivalent to a, b, c being multiples of 5. The symmetric sums are σ₁ = −a, σ₂ = b, and σ₃ = −c, so all three must be divisible by 5. Since p(n) ≡ 0 (mod 5) for n = 1, 2, 3, 4, the polynomial is divisible by (x − 1)(x − 2)(x − 3)(x − 4) modulo 5. I am computing that product mod 5. The product (x − 1)(x − 2)(x − 3)(x − 4) equals x⁴ − 1 modulo 5. The constant terms match: (−1)(−2)(−3)(−4) = 24 ≡ 4 ≡ −1 (mod 5). The polynomial is exactly x⁴ − 1 modulo 5 when a, b, c are multiples of 5. Because the roots are integers, their multiset modulo 5 must be {1, 2, 3, 4}. The roots are of the form 5k+1, 5m+2, 5n+3, and 5p+4 for integers k, m, n, p. Their product is −21. Because the absolute value of each root must divide 21, the possibilities are limited. The absolute values of the roots must be divisors of 21, so each is 1, 3, 7, or 21. Because the product is negative, an odd number of roots are negative. The only possible multisets of absolute values are {1, 1, 1, 21} and {1, 1, 3, 7}. Other permutations of these sets are equivalent. A root of −21 is impossible because the product must be −21. The divisors of 21 modulo 5 are exactly 1, 2, 3, and 4. The four roots must therefore be a permutation of one root from each residue class. There are sixteen possible combinations of one root from each residue class. I will check which product equals −21. The absolute values of the roots are 1, 3, 7, and 21. Because the product of the absolute values is 21, one pair must be 3 and 7 while the other pair is 1 and 21. This fixes the possible combinations for the four roots. The only valid combination is one root from each residue class with absolute values 1, 3, 7, and 21. The first case gives roots 1, −3, −7, and −1, whose product is −21. The second case with roots 1, 7, 3, and −1 is also being checked. The second case also yields a product of −21. No other combinations of absolute values are possible, so the two valid sets of roots are 1, −3, −7, −1 and 1, 7, 3, −1. A root of 21 was considered, but it would force the other three roots to multiply to −1. That would duplicate the residue class 1 modulo 5 and fail to cover all four nonzero classes. The earlier conclusion that the roots must be a permutation of one element from each residue class remains valid. The first set of roots gives the polynomial x⁴ + 10x³ + 20x² − 10x − 21. The second set will be checked next. The second set of roots yields the polynomial x⁴ − 10x³ + 20x² + 10x − 21. Both sets therefore give a² + b² + c² = 600, which is option B. The thought ends by questioning whether any other possibilities remain. The two roots with absolute value 1 must be 1 and −1. Only these can provide the residue classes 1 and 4 modulo 5, because 3 and 7 are not congruent to 1 or 4. This fixes the remaining roots as one element from each of the classes 2 and 3. The thought briefly considers roots 21, 1, 1, −1. This set fails because the product is −21 but the modulo classes are not distinct. It also checks that a = −22 is not a multiple of 5. The thought briefly considers roots 21, 1, 1, −1. This set fails because the product is −21 but the modulo classes are not distinct. It also checks that a = −22 is not a multiple of 5. The case with roots 21, 1, 1, −1 is ruled out because p(2) would not be zero modulo 5. The remaining candidate set {1, 1, 3, 7} still needs to be checked by testing all sign combinations that keep the product negative. The two valid root sets are the only ones that satisfy the modulo condition. The remaining sign combinations for the absolute values 1, 3, 7, and 21 either fail to cover all nonzero residue classes or do not yield a multiple of 5 for the symmetric sums. Therefore no other solutions exist.

We need answer problem in Portuguese. Need solve. Let's restate. Polynomial p(x)=x^4 + a x^3 + b x^2 + c x -21, with a,b,c reals. Knows p(x) has four integer roots and for every integer n not multiple of 5, p(n) is multiple of 5. Determine a^2+b^2+c^2. Options: 375,600,1368,1624,2025. We need solve. Let's analyze. Given monic quartic with integer? It has four integer roots. Let roots r1,r2,r3,r4 integers. Product roots = -21 (constant term -21 = (-1)^4 product? For monic degree 4: constant = product roots = -21). Yes product r1 r2 r3 r4 = -21. Since integer roots, each root divisor of -21? Wait product = -21, each root integer, possible divisors ±1,±3,±7,±21. Product -21 means one negative? Actually product negative, so odd number of negative roots. Possibilities: four roots integers with product -21. Also condition: for every integer n not divisible by 5, p(n) ≡0 mod5 (multiple of 5). Since p has integer coefficients? a,b,c reals? Wait coefficients reals but polynomial with integer roots maybe symmetric? But roots integers imply coefficients? For monic polynomial with integer roots, coefficients are integer? Elementary symmetric sums of integer roots are integers, so a,b,c integers actually. Yes. Condition p(n) ≡ 0 mod 5 for all n not multiple of 5. Since p(x) mod5? If coefficients integer, p(x) mod5 is polynomial over F5 of degree ≤4. It vanishes on all nonzero residues mod5? F5 has 4 nonzero elements. Degree ≤4 polynomial with 4 roots? Over field, degree 4 polynomial can have at most 4 roots unless zero polynomial. Since p vanishes at 4 nonzero elements (1,2,3,4 mod5), then p(x) mod5 must be zero polynomial (all coefficients divisible by 5) because degree ≤4 and four roots? Wait if degree exactly 4, 4 roots possible, but could be scalar times (x-1)(x-2)(x-3)(x-4) = x^4 -1 over F5? Since (x-1)(x-2)(x-3)(x-4)=x^4-1 mod5? Let's compute: In F5, roots all nonzero: 1,2,3,4=-1. Product (x-1)(x-2)(x-3)(x-4) = (x^2-1)(x^2-4?) Actually (x-1)(x-4)=x^2-5x+4=x^2-1 mod5; (x-2)(x-3)=x^2-5x+6=x^2+1. Product = (x^2-1)(x^2+1)=x^4-1 mod5. Yes. So a nonzero degree 4 polynomial vanishing on all nonzero elements is c(x^4-1) with c≠0 mod5. But p has constant -21 ≡ -1 ≡4 mod5? Wait -21 mod5 = -1 ≡4. If p(x) ≡ λ(x^4-1) mod5, constant term ≡ -λ ≡4 => λ≡1? Actually -λ ≡4 => λ≡1 mod5? Since -1≡4, λ=1 gives -1≡4. So possible p(x) ≡ x^4 -1 mod5, i.e., a,b,c ≡0 mod5? Compare x^4 + a x^3 + b x^2 + c x -21 ≡ x^4 + 0 x^3 +0 x^2 +0 x +4 mod5. So a≡b≡c≡0 mod5, and constant -21≡4 yes. Thus a,b,c multiples of 5. Good. But wait condition for every integer n not multiple of 5. It includes n ≡1,2,3,4 mod5. So yes p mod5 vanishes on all four nonzero residues. Since degree ≤4, p mod5 is scalar multiple of product ∏_{r=1}^4 (x-r). As above. So a,b,c divisible by 5. Let a=5A, b=5B, c=5C integers. Need find possible roots / a^2+b^2+c^2. Polynomial p(x)=x^4 + 5A x^3 +5B x^2 +5C x -21. Four integer roots. Product roots = -21. Sum roots = -a = -5A => multiple of 5? Sum roots ≡0 mod5. Pair sum = b =5B => multiple of 5. Sum triple = -c = -5C => multiple of 5. So all elementary symmetric sums of roots except product are multiples of 5. Product = -21 ≡4 mod5. Let roots r_i integers. Need find multiset of four integers product -21 and elementary sums conditions? Actually a,b,c multiples of 5 means sum roots ≡0 mod5, sum pairs ≡0 mod5, sum triples ≡0 mod5 mod5. Since p mod5 = x^4 -1 = (x-1)(x-2)(x-3)(x-4) = x^4 -1? Wait product roots modulo 5? If roots modulo 5 are all nonzero distinct? Because p mod5 = x^4 -1 = ∏_{r=1}^4 (x-r) over F5. Therefore roots modulo 5 must be exactly 1,2,3,4 (multiset) in F5? Since factorization unique; each root must be congruent to one of 1,2,3,4 modulo 5 and none divisible by 5. Also product roots ≡1*2*3*4 =24≡4 mod5 = -21 mod5 yes. Since product = -21 ≡4. So each root r_i is not divisible by 5, and their residues modulo 5 are 1,2,3,4 in some order (since four roots and four nonzero residues, each exactly once? Because product factorization degree 4; yes roots mod5 must be all four nonzero residues, each root congruent to a distinct nonzero residue). Thus {r_i mod5} = {1,2,3,4} as multiset. Also product of actual roots = -21. Now integer roots with product -21. Possible roots: divisors ±1,±3,±7,±21. Need four integers (not necessarily distinct? Yes roots may repeat? But polynomial has four integer roots; could have multiple roots. However if repeated, modulo factorization would have repeated roots? But p mod5 has distinct roots 1,2,3,4, no repeats. Therefore roots modulo 5 distinct, so actual roots must be distinct modulo 5; hence no two roots congruent mod5. So four distinct roots among divisors, each residue 1,2,3,4 mod5. Let's list divisors of -21: 1,3,7,21,-1,-3,-7,-21. Residues mod5: 1→1 3→3 7→2 21→1 (since 21≡1) -1→4 -3→2 (because -3≡2) -7→3 ( -7≡3? -7+10=3) -21→4 ( -21+25=4) We need choose four roots with product -21, residues 1,2,3,4 distinct. Also each root not divisible by 5 automatically except? None. Let's enumerate possible subsets of 4 divisors with product -21 and residues distinct. Divisors: +1 (res1), +3 (res3), +7 (res2), +21 (res1), -1 (res4), -3 (res2), -7 (res3), -21 (res4). Product of four roots = -21. Since product negative, number of negative roots odd (1 or 3). Case 1 negative: one negative root, three positive roots. Product = (negative)*(positive product)= -21 => positive product =21. Positive divisors product for three numbers =21. Possible positive triples with product 21: 1,1,21? But roots distinct mod5? residues: 1,1,1 not distinct (need one each? But if three positive and one negative, residues among positives must cover remaining residues after negative). Product of positive =21, possible triples: (1,1,21) product21; (1,3,7) product21. But roots distinct? (1,3,7) residues 1,3,2. Then negative must have residue 4 to complete {1,2,3,4}. Negative divisor with residue4: -1 or -21. Product total: if positive (1,3,7) product21, negative -1 => product -21; negative -21 => product -441 not -21. So negative must be -1. Roots {1,3,7,-1}? But product 1*3*7*(-1)=-21 yes. Residues: 1->1, 3->3, 7->2, -1->4. All distinct. So multiset {1,3,7,-1} works? Check polynomial roots? Sum=10? 1+3+7-1=10 => a=-10 not divisible by5? Wait a=-sum = -10 divisible by5? -10 ≡0 mod5 yes. b=sum pairs? 1*3+1*7+1*(-1)+3*7+3*(-1)+7*(-1)=3+7-1+21-3-7=20? 3+7=10-1=9+21=30-3=27-7=20 divisible by5. c=-sum triples = -(1*3*7 +1*3*(-1)+1*7*(-1)+3*7*(-1)) = -(21 -3 -7 -21)= -(-10)=10 divisible by5. Constant -21 yes. So possible a=-10, b=20, c=10 => a^2+b^2+c^2=100+400+100=600. Option B. Is that unique? Need check other possibilities. Case 1 negative product positive 21: (1,1,21) product 21. Residues 1,1,1 not distinct; need residues {1,2,3,4} with negative residue4. If positive triple (1,1,21) residues 1,1,1, missing 2,3, can't with one negative. So no. Case 3 negatives: three negative roots, one positive. Product total = -21 => positive * product(negatives)= -21 => product(negatives)= -21? Actually three negative product negative, times positive positive => negative = -21 => product of absolute values =21? Let negative roots -a,-b,-c with a,b,c positive product abc=21? Because (-a)(-b)(-c)= -abc; times positive d = -abcd = -21 => abcd=21. So positive root d and positive abs of negatives multiply 21. Four positive integers product 21. Possibilities: {1,1,1,21}, {1,1,3,7} (unordered). Residues of actual roots: positive d residue d mod5; negatives -a etc residues = -a mod5. Need all four residues {1,2,3,4}. Try multiset actual maybe {-1,-1,-1,21}? Product (-1)^3*21=-21. Residues: -1→4 (three times), 21→1. Missing 2,3; no. Try { -1,-1,-3,7}? Product (-1)(-1)(-3)(7)= -21? (-1*-1*-3)= -3; *7=-21 yes. Residues: -1→4, -1→4, -3→2, 7→2. Missing 1,3; no. Try {-1,-1,-7,3}? Product (-1*-1*-7)*3 = -21? (-1*-1=1; *-7=-7; *3=-21 yes). Residues: -1→4 twice, -7→3, 3→3: missing 1,2; no. Try {-1,-3,-7,1}? Product (-1*-3*-7)*1 = -21? (-1*-3=3; *-7=-21; *1=-21 yes). Residues: -1→4, -3→2, -7→3, 1→1. All distinct {1,2,3,4}. So roots {-1,-3,-7,1} works? But same as positive? Actually multiset positive {1,3,7,-1} we already had. Yes. Try {-1,-1,-21,1}? Product (-1*-1*-21)*1 = -21? (-1*-1=1; *-21=-21; *1=-21). Residues: -1→4 twice, -21→4 (since -21≡4), 1→1: missing 2,3; no. Try {-1,-3,-1,7}? same as -1,-1,-3,7 checked. Try {-3,-7,-1,1}? same. Positive product 21 possibilities: 1,3,7 and negative -1; or 1,1,21 and negative? Need product positives 21 => 1*1*21=21 but negative -1 gives total -21? Actually if three positives 1,1,21 and one negative -1 product -21. Roots {1,1,21,-1}. Residues: 1→1 twice, 21→1 (since 21≡1), -1→4. Missing 2,3; no. If negative -? product of positives 21, negative must -1 for total -21. So no. What about less than 4 roots? "possui quatro raízes inteiras" means degree 4 and all roots integer? Usually means all roots (counting multiplicity) are integers maybe. Could be repeated. We considered multiplicities maybe not explicitly but residues distinct prevents repeats? Wait if root repeated, modulo 5 repeated? But p mod5 = x^4 -1 has no repeated roots because derivative 4x^3 ≠0 mod5 since 4≠0 mod5 and x≠0. So no repeated roots modulo 5. Thus actual roots cannot be congruent mod5, but could be equal? If two roots equal same integer r, then residues same, impossible because then p has repeated root mod5. So roots distinct. Good. Case maybe one negative root not necessarily -1? We assumed positive product 21. If one negative root n, positive roots a,b,c product abc=21. Positive divisors product 21 only (1,1,21) or (1,3,7). Because positive divisors. Negative n must product n*abc=-21 => n=-1. So only -1. Yes. Case three negatives product abs 21: negative roots -a,-b,-c with abc=21, positive d=1? Actually positive root d * abc =21, so d and abc product 21. Since four positive integers, possibilities {1,1,1,21}, {1,1,3,7}. But d can be any divisor? Need residues cover all. We checked. But wait could two negative roots and two positive roots? Product positive? No product -21 requires odd number negatives. So only 1 or 3 negatives. Thus only one multiset {1,3,7,-1}. Then a^2+b^2+c^2=600. But need ensure no other a,b,c due to order? Roots multiset unique so polynomial unique? Let's compute polynomial from roots {1,3,7,-1}: (x-1)(x-3)(x-7)(x+1) =? Let's expand: (x-1)(x+1)=x^2-1; (x-3)(x-7)=x^2-10x+21; product = (x^2-1)(x^2-10x+21)=x^4 -10x^3 +21x^2 - x^2 +10x -21? Wait compute: x^2*(x^2-10x+21)=x^4-10x^3+21x^2; -1*(x^2-10x+21)= -x^2+10x-21; sum = x^4 -10x^3 +20x^2 +10x -21. So a=-10, b=20, c=10 => sum squares 600. Yes. But let's double-check condition: For every integer n not multiple of 5, p(n) multiple of 5. Our polynomial p(x)=x^4 -10x^3 +20x^2 +10x -21 ≡ x^4 -0 -0 +0 -1 ≡ x^4 -1 mod5? Actually 20x^2≡0, 10x≡0, -21≡ -1 ≡4. Yes x^4 -1. For n not multiple 5, n^4 ≡1 mod5 by Fermat, so p(n)≡0. Good. But wait, condition says for every integer n that is not multiple of 5, p(n) is multiple of 5. This is equivalent p(x) mod5 vanishing on all nonzero residues. We used. But is it possible p(x) mod5 = zero polynomial? Degree 4 poly zero mod5 would require all coefficients divisible by5 including leading coefficient 1 not divisible => impossible. So yes p mod5 = λ(x^4-1). Since leading coefficient 1, λ=1. So a,b,c multiples of 5 and constant -21 ≡ -1 mod5 =4. Good. However, need be careful: The polynomial has four integer roots but maybe not necessarily all roots are integers? Wording "possui quatro raízes inteiras" means it has four integer roots (all roots are integers) likely. If it had repeated roots? Degree 4, four integer roots (counting multiplicity). Yes. Need prove uniqueness rigorously. Let's devise elegant solution. Given p(x)=x^4 + a x^3 + b x^2 + c x -21 with integer roots r_i. Product roots = -21. All roots not divisible by 5? Actually if a root divisible by 5, then p(5k)=? Condition n=5k? Not applicable because n multiple of5 excluded. But p(0)? n=0 multiple of5? Condition only n not multiple of5; 0 is multiple of5, excluded. But p(r)=0 for root r; if r divisible by5? Then condition doesn't force p(r) multiple of5 because r multiple of5. But polynomial degree? But product roots -21 not divisible by5? If one root divisible by5, product divisible by5, contradiction -21 not divisible by5? Wait -21 not divisible by5. Product roots = -21 not divisible by5, so no root divisible by5 (since product would be divisible by5 if any root divisible by5? Actually if one root divisible by5, product divisible by5, but -21 not divisible by5, contradiction). So roots not divisible by5. Good. Then modulo 5, p(x) = ∏(x - r_i mod5). Since p mod5 vanishes at all nonzero residues, each nonzero residue must be a root of p mod5. Since degree 4 and there are 4 nonzero residues, the roots modulo 5 must be exactly the four nonzero residues (counting multiplicity). Therefore {r_i mod5} = {1,2,3,4}. In particular the residues are distinct, so roots distinct mod5. Now list divisors of -21 and residues. Need choose four with product -21 and residues 1,2,3,4. Could enumerate. Alternative: Since product roots = -21, possible roots are divisors. Let's list all divisors with residues: 1 (1), 3 (3), 7 (2), 21 (1), -1 (4), -3 (2), -7 (3), -21 (4). Need pick four product -21. Since product negative, odd negatives => either 1 or 3 negatives. If 1 negative: three positives product 21. Possible positive triples product 21: {1,1,21} or {1,3,7}. But residues of positives must be three distinct among {1,2,3} (since negative will provide 4). {1,1,21} residues 1,1,1 not distinct; {1,3,7} residues 1,3,2 distinct. Negative must have residue 4: options -1 or -21. But product with -21 would be 1*3*7*(-21)= -441 ≠ -21. So negative must be -1. Thus roots {1,3,7,-1}. If 3 negatives: absolute values product 21 with one positive. Four absolute values product 21. Possibilities: {1,1,1,21} and {1,1,3,7}. Assign signs: three negatives among four abs values, positive one. Need residues of actual roots = {1,2,3,4}. Let's test possibilities. For abs {1,1,1,21}: possible assignments: - Three 1's negative and 21 positive: roots {-1,-1,-1,21}. Residues {4,4,4,1} not all. - Two 1 negative, one 1 positive, 21 negative: {-1,-1,1,-21} residues {4,4,1,4} not all. - One 1 negative, two 1 positive? But only three negatives total; if two positives among 1's and one positive 21? Wait four numbers: three negatives and one positive. The positive could be one of the 1's or 21. Cases: Positive=21, negatives -1,-1,-1 => residues 4,4,4,1 no. Positive=1, negatives -1,-1,-21 => residues 4,4,4,1? -21≡4, 1≡1, missing 2,3 no. Positive=1, negatives -1,-21,-1 same. No. For abs {1,1,3,7}: possible assignments with three negatives: Let's enumerate all sign patterns with exactly three negatives (odd). Positive could be one of 1,1,3,7; the other three negative. - Positive 1: roots {-1,-1,-3,-7}? Actual numbers: abs 1 negative -1, abs 1 negative -1, abs 3 negative -3, abs 7 negative -7 => {-1,-1,-3,-7}. Residues: 4,4,2,3 missing 1. - Positive other 1 same. - Positive 3: roots {-1,-1,3,-7}? Abs 1 neg -1, abs 1 neg -1, abs 3 positive 3, abs 7 neg -7 => residues 4,4,3,3 missing 1,2. - Positive 7: roots {-1,-1,-3,7} => residues 4,4,2,2? 7≡2, -3≡2, missing 1,3. - Positive? What about abs 21 not in set. So no. Thus unique roots. Then compute a,b,c. Alternative using symmetric sums: Roots 1,3,7,-1. Sum =10 => a=-10. Pair sum = b = 1*3+1*7+1*(-1)+3*7+3*(-1)+7*(-1)=3+7-1+21-3-7=20. Triple sum = -c = 1*3*7 +1*3*(-1)+1*7*(-1)+3*7*(-1)=21-3-7-21=-10 => c=10. a^2+b^2+c^2=100+400+100=600. Answer B. Need ensure no other polynomials? Let's consider maybe polynomial with complex roots? But statement "possui quatro raízes inteiras" means all four roots are integers (since degree 4). Yes. Potential issue: The condition "para todo inteiro n que não seja múltiplo de 5, o valor de p(n) é múltiplo de 5". Does "múltiplo de 5" include negative multiples? Yes. Maybe there is a catch: n not multiple of 5 includes n=±1,±2,...; p(n) multiple of 5. Since p has integer coefficients, p mod5 zero on nonzero residues. We concluded p mod5 = x^4-1. But wait, if p mod5 is degree less than 4? Leading term 1 so degree 4. Over F5, a degree 4 polynomial can vanish on 4 points without being zero; indeed possible scalar times product. So a,b,c divisible by5? Compare coefficients: p mod5 = x^4 + a x^3 + b x^2 + c x + (-21 mod5) = x^4 + a x^3 + b x^2 + c x +4. For this to equal x^4 -1 = x^4 +4 (since -1=4), coefficients of x^3,x^2,x must be 0 mod5: a≡0, b≡0, c≡0 mod5. Yes. But wait, what if p mod5 = 2(x^4-1)? Leading coefficient would be 2, but p leading coefficient is 1 mod5 (1≠0). So no. So exactly. Need maybe prove product roots = -21? For monic polynomial x^4 + a x^3 + b x^2 + c x -21 = ∏(x - r_i) = x^4 - (sum r_i)x^3 + (sum pairs)x^2 - (sum triples)x + product r_i. Thus product r_i = -21. Yes. Let's craft final explanation in Portuguese. Need be concise but complete. Include steps: 1. Since roots integers and product -21, possible roots divisors. 2. Condition modulo 5 implies p(x) ≡ x^4 -1 mod5, so roots modulo 5 are exactly 1,2,3,4; hence a,b,c multiples of 5. 3. Enumerate possible roots using product and residues. 4. Compute polynomial and sum squares. Let's write carefully. Potential alternative: Could derive using factorization modulo 5 directly: p(x) ≡ (x-1)(x-2)(x-3)(x-4) mod5 = x^4 -1. Then a,b,c divisible by5. Also p(x) has integer roots; modulo 5, roots of p are 1,2,3,4. Then use integer roots divisors. Let's ensure enumeration. Residues: r=1 =>1 r=3 =>3 r=7 =>2 r=21 =>1 r=-1 =>4 r=-3 =>2 r=-7 =>3 r=-21 =>4 Need choose four roots with product -21. Since product negative, odd number of negative roots. If only one negative root n, then product of three positive roots =21. Positive triples from divisors: (1,1,21) or (1,3,7). But residues of positives must be three distinct residues from {1,2,3} (because negative provides 4). (1,1,21) residues {1,1,1} not distinct; (1,3,7) residues {1,3,2}. Thus positive roots are 1,3,7. Negative root must have residue 4 and product with positives -21 => n=-1 because n*21=-21 => n=-1. So roots {1,3,7,-1}. If three negative roots, product of their absolute values =21 (since positive root times (-1)^3*product abs = -21 => positive * product abs =21). So four positive absolute values product 21: {1,1,1,21} or {1,1,3,7}. Assign signs to get three negatives and one positive. Check residues cannot cover {1,2,3,4}; we can state "verifica-se facilmente" or show. Let's detail: - Absolute values {1,1,1,21}: any sign assignment with three negatives leaves at least three roots congruent to 4 (-1) or 1? Let's list possible actual multisets and residues: * {-1,-1,-1,21}: residues 4,4,4,1 -> missing 2,3. * {-1,-1,1,-21}: residues 4,4,1,4 -> missing 2,3. Other assignments with positive 1? Actually there are C(4,3)=4 sign assignments? Since three negatives among four numbers; but two numbers 1 identical. The distinct actual multisets: choose which one positive. If positive 21, negatives three 1 => {-1,-1,-1,21}. If positive one 1, negatives: the other 1 and 21 and one? Wait there are three 1's and one 21. If positive is one 1, then negatives are two 1's and 21 => {-1,-1,1,-21}? Wait positive 1, negatives -1,-1,-21 => multiset {-1,-1,1,-21}. Residues 4,4,1,4. If positive 21 same as first. So only two distinct multisets, no. - Absolute values {1,1,3,7}: possible distinct multisets: Positive 1: negatives two 1s? Actually abs values: two 1s, 3,7. If positive one 1: negatives -1, -3, -7 => {-1,-1,-3,-7} residues 4,4,2,3 missing 1. Positive other 1: same. Positive 3: negatives -1,-1,-7 => {-1,-1,3,-7} residues 4,4,3,3 missing 1,2. Positive 7: negatives -1,-1,-3 => {-1,-1,-3,7} residues 4,4,2,2 missing 1,3. Positive? Could positive be 3 and negatives include -3? Wait abs 3 can be negative -3 or positive 3. Yes. What about positive = 1, negatives include -7 and -3 and -1: done. No all residues. Thus unique. Then compute. Let's ensure no issue with roots order? No. Answer B. Let's produce final in Portuguese, with explanation. Desired oververbosity 5, so moderate. Keep final answer line exactly "Resposta: B". Need maybe mention "a^2+b^2+c^2 = 600". Let's craft final solution. Potential nuance: The polynomial could have roots not necessarily distinct? But we argued distinct mod5. However, our enumeration assumed four roots distinct? Actually if repeated roots, product? But modulo 5 distinct so no repeats. Good. Let's write: "Como p tem coeficientes inteiros? Na verdade, como todas as raízes são inteiras, os coeficientes a,b,c são inteiros (somas simétricas)." Need justify? If roots integers, coefficients are integers by Vieta. So yes. Then "p(x) ≡ x^4 -1 mod5" derivation: For n=1,2,3,4 mod5, p(n)≡0. O polinômio q(x)=p(x) mod5 tem grau 4 e se anula nos quatro elementos não nulos de F5. Logo q(x)=λ(x-1)(x-2)(x-3)(x-4)=λ(x^4-1). Comparando coeficientes líderes, λ=1. Portanto p(x)≡x^4-1 mod5, donde a≡b≡c≡0 mod5. Actually "coeficientes líderes" both 1. Yes. Then "Assim, as raízes inteiras r_i devem ter resíduos 1,2,3,4 módulo 5, pois p mod5 = ∏(x-r_i) = x^4-1 = ∏_{k=1}^4 (x-k)." Then enumeration. Let's final.

**Analyzing Polynomial Roots** I'm currently focused on deciphering the integer roots of the polynomial $p(x) = x^4 + ax^3 + bx^2 + cx - 21$. The key constraint is that for any integer $n$ not divisible by 5, $p(n)$ is a multiple of some value, which implies a relationship with the roots. **Exploring Root Congruences** I'm now investigating how the polynomial's roots behave modulo 5. Specifically, I'm using the condition that $p(n)$ is a multiple of 5 for all $n$ not divisible by 5 to constrain the possible values of these integer roots within $\mathbb{Z}_5$. **Deriving Root Modulo Congruences** I've confirmed that the four roots of the polynomial, when considered modulo 5, must form a complete residue system $\{1, 2, 3, 4\}$. This is because for any integer $n$ not divisible by 5, $p(n)$ must be divisible by 5, forcing each root to be congruent to one of $\{1, 2, 3, 4\}$ modulo 5. **Testing Root Combinations** I've established that the roots modulo 5 are $\{1, 2, 3, 4\}$, which aligns with the constant term modulo 5. I'm now exploring combinations of integer roots whose product is $-21$, focusing on absolute value sets $\{1, 1, 3, 7\}$ and $\{1, 1, 1, 21\}$. **Refining Root Congruences** I've eliminated possibilities for root congruences modulo 5. It appears that roots must be congruent to 1 or 4 (i.e., $\pm 1$) modulo 5, ruling out options involving 2 or 3. The absolute values of the roots are likely $\{1, 1, 3, 7\}$. **Testing Root Scenarios** I've determined that two roots must be $+1$ and $-1$ modulo 5 to satisfy the conditions. This means their product is $-1$, leaving the product of the remaining two roots to be $21$. They must therefore have the same sign. **Verifying Root Pairs** I've confirmed that my derived root congruences modulo 5, $\{1, 4\}$ for two roots and implicitly $\{2, 3\}$ for the other two, are consistent with the constant term. Now, I'm examining potential integer root pairs whose product is 21, specifically $\{1, -1, -3, -7\}$, which aligns with these modular constraints. **Exploring Root Sign Choices** I have identified two potential sets of integer roots: $\{1, -1, 3, 7\}$ and $\{1, -1, -3, -7\}$. The latter set's roots are the negations of the former, leading me to investigate the relationship $p_B(x) = p_A(-x)$. **Analyzing Polynomial Symmetry** I've established that the polynomial with roots $\{1, -1, -3, -7\}$ is indeed $p_A(-x)$ where $p_A(x)$ has roots $\{1, -1, 3, 7\}$. This implies a relationship between the coefficients $a, c$ and the polynomial itself, which I'm now exploring. **Calculating Polynomial Coefficients** I have successfully determined the coefficients for one of the potential root sets. This involved expanding the factored form of the polynomial and verifying the sums and products of the roots against the derived coefficients. The uniqueness of $a^2 + b^2 + c^2$ is now confirmed. **Confirming Final Calculation** I have verified the calculation for $a^2 + b^2 + c^2$. The sum is indeed 600, aligning perfectly with one of the provided options. I am now meticulously reviewing all prior steps and constraints to ensure absolute certainty in this result. **Refining Root Modulo 5** I've confirmed the roots modulo 5 must be $\{1, 2, 3, 4\}$. With the absolute values of the integer roots being either $\{1, 1, 3, 7\}$ or $\{1, 1, 1, 21\}$, the latter is ruled out as it cannot yield four distinct residues modulo 5.