MHB Can You Solve the Triple Integers System from POTW #122?

[anemone]

Find all triples $(a,\,b,\,c)$ of positive integers satisfying the system of equations

$a^2=2(b+c)$

$a^6=b^6+c^6+31(b^2+c^2)$.

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[anemone]

Congratulations to Opalg for his correct solution, as shown below:

Spoiler

First, notice that $(a,b,c) = (2,1,1)$ is a solution.

The problem is symmetric in $b$ and $c$, so we may as well assume that $b\leqslant c.$ Then $a^2 = 2(b+c) \leqslant 4c$. Therefore $$(c^2)^3 = c^6 < b^6 + c^6 + 31(b^2+c^2) = a^6 \leqslant (4c)^3.$$ It follows that $c^2 < 4c,$ so that $c<4.$ Thus the only possible values for $b$ and $c$ are $1$, $2$ or $3$. But $a^2 = 2(b+c)$, and the only pair of numbers between $1$ and $3$ for which twice their sum is a square is $b=c=1$.

Therefore the only solution to the problem is $(a,b,c) = (2,1,1).$