$p$ or $q$

$p+q=(p-q)^3$

$p>q$ , therefore p is not the smallest prime, and $p$ is odd

let $k$ be any odd factor of $p-q$ .

therefore $k$ is also a factor of $p+q$

therefore $k$ is a factor of $2q$ , the difference.

But $2q$ has NO odd proper factors, so k does not exist, and therefore $p-q$ has no odd factors.

(considering the case $k=q$ , k cannot then be a factor of $p+q$ because $p$ is prime)

therefore p-q is a power of 2, say 2^m

$p-q=2^m$

$p+q=2^{3m}$

$2^{3m}-2^m=2q$

$2^{3m-1}-2^{m-1}=q$ and is therefore odd

therefore $2^{m-1}$ is odd

therefore $m=1$

$q=3$ , $p=5$

$pq=15$

We do the substitution $n=p-q$ , note that $n$ will always be positive because for the equation to be true p will always be greater than q.

$2p-n=n^3$

$p=\frac{1}{2}n(n^2+1)$

Already my intuition tells me that $n=2$ is an answer. But since intuition is not rigorous, we shall continue:

Since p is prime one of the following should be true:

• $n=1$ => $p=1$ and $q=0$ . That is wrong.
• $\frac{n}{2}=1$ => $n=2$ => $p=5$ and $q=3$ . This is an answer to the equation.
• $n^2+1=1$ => $n=0$ => $p=q=0$ that is wrong
• $\frac{n^2+1}{2}=1$ => (n=1) something that we know to be wrong.

In conclusion: p=5 and q=3 is the only solution to the equation.