Let n be a positive integer such that n5+n3+2n2+2n+2 is a perfect cube. Prove that 2n2+n+2 is not a perfect cube. Proposed by Anastasija Trajanova
Problem
Source: 2022 Junior Macedonian Mathematical Olympiad P5
Tags: polynomial, number theory
08.06.2022 17:18
Revised and simplified version: Let P(n)=n5+n3+2n2+2n+2=p3,Q(n)=2n2+n+2=q3 ⟹P(n)−Q(n)=n5+n3+n=p3−q3 Taking mod 9⟹n5+n3+n mod 9∈ {0,3,6} (see below) p3,q3 mod 9∈ {0,1,8}⟹p3−q3 mod 9∈ {0,1,2,7,8} Therefore, the only acceptable option is n=9k for the LHS and p=9l,q=9r or p=9l+i,q=9r+i for the RHS However, replacing n=9k in n5+n3+n=p3−q3⟹95k5+93k3+9k=93l3−93r3 (or =(9l+i)3−(9r+i)3) shows that the RHS is divisible by 93 or 33, unlike the LHS. Thus also this option fails and no n satisfies the given requirements. n=9k+1⟶n5=1,n3=1,n=1,n5+n3+n≡3mod n=9k+2\longrightarrow n^5=5, n^3=8, n=2, n^5+n^3+n\equiv 6\bmod{9} n=9k+3\longrightarrow n^5=0, n^3=0, n=3, n^5+n^3+n\equiv 3\bmod{9} n=9k+4\longrightarrow n^5=7, n^3=1, n=4, n^5+n^3+n\equiv 3\bmod{9} n=9k+5\longrightarrow n^5=2, n^3=8, n=5, n^5+n^3+n\equiv 6\bmod{9} n=9k+6\longrightarrow n^5=0, n^3=0, n=6, n^5+n^3+n\equiv 6\bmod{9} n=9k+7\longrightarrow n^5=4, n^3=1, n=7, n^5+n^3+n\equiv 3\bmod{9} n=9k+8\longrightarrow n^5=8, n^3=8, n=8, n^5+n^3+n\equiv 6\bmod{9} .
08.06.2022 17:26
Assume that n^5 + n^3 + 2n^2 + 2n + 2 = a^3 and 2n^2 + n + 2 = b^3 for positive integers a and b. Then (a - b)(a^2 + ab + b^2) = a^3 - b^3 = n^5 + n^3 + n = n(n^2 - n + 1)(n^2 + n + 1). Note that RHS is divisible by 3, so at least one of a - b or a^2 + ab + b^2 = (a - b)^2 + 3ab has to be divisible by 3. Hence, both a - b and a^2 + ab + b^2 are divisible by 3, and 9 | n(n^2 - n + 1)(n^2 + n + 1). Since n, n^2 - n + 1, and n^2 + n + 1 are pairwise relatively prime, we have three cases: Case 1: 9 | n. Then 2n^2 + n + 2 \equiv 2 \pmod{9}, which cannot be a perfect cube. Case 2: 9 | n^2 - n + 1. Then n^2 - n \equiv 8 \pmod{9} \iff (2n-1)^2 \equiv 6 \pmod{9}, which is impossible, since all perfect squares divisible by 3 are divisible by 9. Case 3: 9 | n^2 + n + 1. Then n^2 + n \equiv 8 \pmod{9} \iff (2n+1)^2 \equiv 6 \pmod{9}, which is impossible, since all perfect squares divisible by 3 are divisible by 9. Thus, n^5 + n^3 + 2n^2 + 2n + 2 and 2n^2 + n + 2 cannot be perfect cubes simultaneously.
08.06.2022 17:54
Note that n is clearly odd, as otherwise n^5+n^3+2n^2+2n+2\equiv 2\pmod{4}. Now, factorize this quantity as (n+1)\left(n^4-n^3+2n^2+2\right); denote by d the gcd of the factors. Then, n\equiv -1\pmod{d} implies n^4-n^3+2n^2+2\equiv 6\pmod{d}, thus d\in\{1,2,3,6\}. As n is odd, d\in\{2,6\}. Now, if 3\mid n+1 then n=3k-1 for some k for which 2n^2+n+2=18k^2-9k+3\equiv 3\pmod{9}, thus it cannot be a perfect cube. Hence, d=2; leaving us with two cases. Case 1. n+1=2a^3. For a=1, 2n^2+n+2=5 is not a cube. Let a>1. Using n=2a^3-1 we get 2n^2+n+2 = 8a^6-6a^3+3, but \left(4a^3-1\right)^2<8a^6-6a^3+3<8a^6, yielding a contradiction. Case 2. n+1=4a^3. In this case, 2n^2+n+2=32a^6-12a^3+3. Note that 3\nmid a, thus a^6\equiv 1\pmod{9} and a^3\equiv \pm 1\pmod{9} (by Euler). Hence, 2n^2+n+2\equiv 35\pm 12 \in\{2,5\}\pmod{9}, none of which is a perfect cube.
08.06.2022 21:02
Assume 2n^2 + n + 2 is a cube. The fact that 2n^2 + n + 2 \equiv 2,5,3,5,2,3,3,8,3 \pmod{9} for n = 0,1,2, \cdots, 8, yields n \equiv 7 \pmod{9} since k^3 \equiv 0,1,8 \pmod{9} for all integers k. Hence 9 \mid n+2, which means n^5 + n^3 + 2n^2 + 2n + 2 \equiv (-2)^5 + (-2)^3 + 2 \cdot (-2)^2 + 2 \cdot (-2) + 2 = -32 - 8 + 8 - 4 + 2 = -34 \equiv 2 \pmod{9}, implying n^5 + n^3 + 2n^2 + 2n + is not a cube. This contradiction completes the proof.
09.06.2022 00:19
Solar Plexsus wrote: Assume 2n^2 + n + 2 is a cube. The fact that 2n^2 + n + 2 \equiv 2,5,3,5,2,3,3,8,3 \pmod{9} for n = 0,1,2, \cdots, 8, yields n \equiv 7 \pmod{9} since k^3 \equiv 0,1,8 \pmod{9} for all integers k. Hence 9 \mid n+2, which means n^5 + n^3 + 2n^2 + 2n + 2 \equiv (-2)^5 + (-2)^3 + 2 \cdot (-2)^2 + 2 \cdot (-2) + 2 = -32 - 8 + 8 - 4 + 2 = -34 \equiv 2 \pmod{9}, implying n^5 + n^3 + 2n^2 + 2n + is not a cube. This contradiction completes the proof. You missed a Case n\equiv 6\mod 9\implies 2n^2+n+2\equiv -1\mod 9 So n\equiv 6\mod 9 is also possible but then n^5+n^3+2n^2+2n+2\equiv 5\mod 9 a contradiction.