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if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime

then $(m,n)=(2,3)$ or $(3,2)$

I found out that one of the $m$ and $n$ must be two since $n^2+m^3$ will be even if $m$ and $n$ have the same parity.

However, how to prove that if $4+n^3$ and $8+n^2$ are both prime, then $n=3$?

yuanming luo
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1 Answers1

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If $n \neq 3$ and is also prime, then $n \equiv 1 \pmod{3} \implies 8 + n^2 \equiv 0 \pmod{3}$, or $n \equiv 2 \pmod{3} \implies 4 + n^3 \equiv 0 \pmod{3}$. Since both $4 + n^3$ and $8 + n^2$ are greater than $3$, this means at least one of them can't be prime.

John Omielan
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  • Thank you!!! I never linked this to modular before. I am still working on how to factorize them... – yuanming luo Dec 21 '20 at 02:35
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    Can simplify: $n^2\equiv 0,1 \bmod 3$. So $8+n^2\equiv 8,9 \equiv -1,0 \bmod 3$. But if $8+n^2\equiv 0 \bmod 3$ it is not prime. This forces $n^2 \equiv 0 \Rightarrow n\equiv 0 \bmod 3$ and the only case in which $n$ is prime is $n=3$. No need to look at $4+n^3$ except to confirm that it is in fact prime. – Keith Backman Dec 21 '20 at 03:27