Ireland_2017

Suppose for the sake of contradiction that $n$ and $z$ are integers satisfying $f(n)=z^3$. Write $f(n)=3(n^4+n+2)+n^4-7n^2$ and let $\tau \in \{0,1,2\}$ be the remainder of $n$ on division by 3.

Suppose 3 divides $z$. Then 3 divides $n^4+7n^2=n^2(n^2+7)$. But $r^2+7\in \{7,8,11\}$ is not divisible by 3. Hence 3 does not divide $n^2+7$ and so 3 must divide $n^2$. This implies that 3 divides $n$ and 9 divides $n^2(n^2+7)$. Since 9 divides $z^3$, it follows that 3 divides $n^4+n+2$. But, with $r$ as before, $r^4+r^2+2$ is not divisible by 3, so 3 does not divide $n^4+n+2$, yielding a contradiction. Suppose that the remainder of $z$ on division by 3 is 1. Then, since 3 divides $f(n)-1$, 3 must divide $n^4+7n^2-1$. Again, $r^4+7r^2-1$ is not divisible by 3, so this case is eliminated. Finally, suppose $z$ leaves remainder 2 on division by 3. Then $z=2+3h$, for some integer $h$, so $z^3=8+9k$, for some integer $k$. Now $n^4+7n^2-8=(n^2-1)(n^2+8)$, and 3 must divide $n^2-1$ or $n^2+8$. But $n^2+8-(n^2-1)=9$, so both factors are divisible by 3, Hence, since $z^3-8$ is divisible by 9 also, 3 must divide $n^4+n+2$. But we have already shown this cannot happen. So we have reached a final contradiction, and the claimed result is established.

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Alternative solution.

Note that the cubes modulo 9 are -1, 0, 1 and observe that $f(n)\equiv 4n^4-2n^2+3n-3=2n^2(2n^2-1)+3(n-1)$ (mod 9). The calculation modulo 9 in the table below shows that $f(n)$ is congruent to -4, -3 or 2 modulo 9, hence cannot be the cube of an integer.

n	0	1	2	3	4	-4	-3	-2	-1
n - 1	-1	0	1	2	3	4	-4	-3	-2
3(n-1)	-3	0	3	-3	0	3	-3	0	3
2n^2	0	2	-1	0	-4	-4	0	-1	2
2n^2(2n^2-1)	0	2	2	0	2	2	0	2	2
f(n)	-3	2	-4	-3	2	-4	-3	2	-4

Therefore, $f(n)$ cannot be the cube of an integer.