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The pairs of primes $(p,q)$ with $p\le q\le 10^5$ such that $p^3+q^3$ is a perfect square , are (the last entry is the maximum possible exponent of the perfect power $p^3+q^3$ which is only different from $2$ in the first example) :

? forprime(a=1,10^5,forprime(b=a,10^5,c=ispower(a^3+b^3);if(c>0,print([a,b,c])))
)
[2, 2, 4]
[11, 37, 2]
[1801, 56999, 2]
[2137, 8663, 2]
[6637, 86291, 2]
[8929, 28703, 2]
[44111, 48817, 2]
[57241, 87959, 2]
?

Are there infinitely many pairs of primes $(p,q)$ such that $p^3+q^3$ is a perfect square?

For fixed $p$, only finitely many $q$ can exist such that the required property holds, but neverhteless infinitely many pairs could exist. Maybe properties of elliptic curves could help solve the question.

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    $\begingroup$ See also math.stackexchange.com/questions/369846/… $\endgroup$
    – lhf
    Commented Jun 13, 2018 at 11:31
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    $\begingroup$ Idle observation: If $p$ and $q$ are both odd and $p^3+q^3$ is a square, then $pq\equiv3$ mod $4$ (i.e., one is congruent to $1$, the other to $3$ mod $4$). $\endgroup$
    – Barry Cipra
    Commented Jun 13, 2018 at 11:59
  • $\begingroup$ oeis.org/A145767 might be of interest. $\endgroup$
    – Barry Cipra
    Commented Jun 13, 2018 at 12:05
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    $\begingroup$ We know the co-primes solutions of $z^2 = x^3 + y^3$ can be parametrized by 3 parametrizations. Only one of them allow $x, y$ to be odd primes: $$x=-3s^4+6t^2s^2+t^4,\quad y=3s^4+6t^2s^2-t^4,\quad z = 6 s t ( 3 s^4+t^4)$$ where $s \not\equiv t \pmod 2$ and $3 \not| t$ (up to exchange of $x,y$). The problem reduces to how many pairs of $(x,y)$ of above form are pair of prime numbers $\endgroup$
    – achille hui
    Commented Jun 13, 2018 at 12:17
  • $\begingroup$ @achillehui Exactly what I need! $\endgroup$
    – Peter
    Commented Jun 13, 2018 at 12:18

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