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Suppose $x$ and $y$ are some integers satisfying $$x^2-16=y^3.$$ I'm trying to show that $x+4$ and $x-4$ are both perfect cubes.

I know that the greatest common divisor of $x+4$ and $x-4$ must divide $8$, but I don't know where to go from there. Would anyone be able to help?

dbanet
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gs2341
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2 Answers2

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We rewrite the equation as $(x-4)(x+4)=y^3$. We know that $\gcd(x-4,x+4) \mid 8$, so for every prime $p>2$ the factors $p$ are either in $x-4$ or $x+4$.

If $y$ is even, at least one of $x-4$ and $x+4$ is divisble by 4, and hence they are both divisible by 4. But then $4 \mid y$, so one of them is divisible by 8 and hence they are both divisible by 8. Then one has a factor 8 and the other has all other factors two.

If $y$ is odd, we don't have any factors 2.

wythagoras
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If prime $p$ divides $x^2-16=(x+4)(x-4),$

$p^3$ must divide $(x+4)(x-4)$

But if $p$ divides both, $p$ must divide $x+4-(x-4)=8$

So, if $p>2,p^3$ must divide exactly one of $x+4,x-4$

For $p=2,$

if the highest power of $2$ that divides $x+4$ is $a$ and $x+4=c2^a$ where $c$ is odd

If $a>3, x-4=8(c2^{a-3}-1),$ then $3$ must divide $a$

If $a<3, x-4=2^a(c-2^{3-a}),$ then $3$ must divide $2a$

If $a=3, x-4=2^a(c-1),$ then $c-1,c$ must be perfect cube which is possible iff $c=0,1$

Hence we are done.

lab bhattacharjee
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