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Find all ordered quadruples $(a, b, c, d)$ of primes such that \begin{align*} 13a^2 + b^2 + 13c^2 &= 2b(3a + 2c) \\ a^3 −(bc)^3 &= (b+c−a)d+2021. \end{align*}

A quick WolframAlpha search yielded that the answer to this was no pairs, but I wasn't quite sure how to derive it.

smythe0173
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1 Answers1

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The first equation, considered as quadratic equation in $c$, has no real solution in $c$, except for the case $3b=13a$.

Then we have $a=3$ and $b=13$, so that the first equation gives $c=2$. This indeed solves the first equation. However, then the second equation is $6d + 9785=0$, so that $d$ is not an integer. Contradiction.

Dietrich Burde
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