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More Prime Congruences

Find all primes p and q such that p+q = (p-q)^3

We first show the trivial property that p \ne q.

Assume by way of deriving a contradiction that p = q, then it must be the case that 2p = 0^3 \iff p,q = 0. This however violates the assumption that p,q are primes. \blackbox

Now, like the previous post, we would like to find some congruence class in order to bound p and q to some class of primes. We would like for the congruence to be derived from the problem statement itself.

    \begin{align*} p + q &= q - p + 2p \\ &= -((p-q) - 2p) \\ &\implies (p+q)|((p-q) - 2p) \\ &\implies p-q \equiv 2p \mod p+q \end{align*}

The last line of the above derivation comes from the definition of congruence as a \equiv b \mod n \iff n|(a-b).

Now, the problem statement is equivalent to

    \begin{align*} (p-q)^3 &\equiv 0 \mod p + q \\ (2p)^3 &\equiv 0 \mod p + q \\ 8p^3 & \equiv 0 \mod p + q \\ &\implies (p+q)|8p^3 \end{align*}

Now, it’s obvious that (p+q,p) = (q,1) = 1 \implies (p+q,p^3) = 1, therefore, if (p+q)|8p^3, and p+q is coprime to p^3, then (p+1)|8. This leaves the following possibilities:

     $$ \begin{tabular}{c | c | c} p+q & (p,q) & (p,q)\\ 1 & \varnothing \\ 2 & \sout{(1,1)} \ding{55} \\ 4 & \sout{(1,3)} \ding{55} & \sout{(2,2)} \ding{55} \\ 8 & (3,5) \cmark \end{tabular} $$

Therefore, only p=5,q=3 satisfies p+q = (p-q)^3. \blacksquare

Posted By Lee

Woosh