Two polynomials cannot be perfect cubes at the same time

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Two polynomials cannot be perfect cubes at the same time

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Source: 2022 Junior Macedonian Mathematical Olympiad P5

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#1 h Jun 7, 2022, 9:11 PM

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Let $n$ be a positive integer such that $n^5+n^3+2n^2+2n+2$ is a perfect cube. Prove that $2n^2+n+2$ is not a perfect cube.

Proposed by Anastasija Trajanova

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#2 h Jun 8, 2022, 2:18 PM • 3 Y

Y by Mango247, Mango247, Mango247

Revised and simplified version:

Let $P(n)=n^5+n^3+2n^2+2n+2=p^3, Q(n)=2n^2+n+2=q^3$

$\Longrightarrow P(n)-Q(n)= n^5+n^3+n=p^3-q^3$

Taking mod $9\Longrightarrow n^5+n^3+n$ mod $9 \in$ { $0, 3,6$ } (see below)

$p^3,q^3$ mod $9 \in$ { $0, 1,8$ } $\Longrightarrow p^3-q^3$ mod $9 \in$ { $0, 1,2, 7, 8$ }

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 $n^5+n^3+n=p^3-q^3\Longrightarrow 9^5k^5+9^3k^3+9k=9^3l^3-9^3r^3$

(or $= (9l+i)^3-(9r+i)^3$ )

shows that the RHS is divisible by $9^3$ or $3^3$ , unlike the LHS. Thus also this option fails and no $n$ satisfies the given requirements.

$n=9k+1\longrightarrow n^5=1, n^3=1, n=1, n^5+n^3+n\equiv 3\bmod{9}$
$n=9k+2\longrightarrow n^5=5, n^3=8, n=2, n^5+n^3+n\equiv 6\bmod{9}$
$n=9k+3\longrightarrow n^5=0, n^3=0, n=3, n^5+n^3+n\equiv 3\bmod{9}$
$n=9k+4\longrightarrow n^5=7, n^3=1, n=4, n^5+n^3+n\equiv 3\bmod{9}$
$n=9k+5\longrightarrow n^5=2, n^3=8, n=5, n^5+n^3+n\equiv 6\bmod{9}$
$n=9k+6\longrightarrow n^5=0, n^3=0, n=6, n^5+n^3+n\equiv 6\bmod{9}$
$n=9k+7\longrightarrow n^5=4, n^3=1, n=7, n^5+n^3+n\equiv 3\bmod{9}$
$n=9k+8\longrightarrow n^5=8, n^3=8, n=8, n^5+n^3+n\equiv 6\bmod{9}$
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This post has been edited 2 times. Last edited by iniffur, Jun 8, 2022, 10:39 PM

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#3 h Jun 8, 2022, 2:26 PM • 1 Y

Y by Jalil_Huseynov

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.

This post has been edited 3 times. Last edited by sarjinius, Jun 8, 2022, 2:37 PM

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#4 h Jun 8, 2022, 2:54 PM • 1 Y

Y by Mango247

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.

This post has been edited 1 time. Last edited by grupyorum, Jun 8, 2022, 4:00 PM

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#5 h Jun 8, 2022, 6:02 PM • 4 Y

Y by grupyorum, Mango247, Mango247, Mango247

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.

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#6 h Jun 8, 2022, 9:19 PM

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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.

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