Let $(a, b, c)$ be a potential solution. We observe that:

$a^{2}<a^{2}+3 b \leq a^{2}+3 a<a^{2}+4 a+4=(a+2)^{2}$ and $b^{2}<b^{2}+3 c \leq b^{2}+3 b<b^{2}+4 b+4=(b+2)^{2}$

Necessarily, $a^{2}+3 b=(a+1)^{2}=a^{2}+2 a+1$ and $b^{2}+3 c=(b+1)^{2}=b^{2}+2 b+1$ so $2 a+1=3 b$ and $2 b+1=3 c$. By combining, we get $a=\frac{9 c-5}{4}$ and we know that $c^{2}+3 a=c^{2}+\frac{27 c-15}{4}$ is a perfect square.

But we know that:

$c^{2}<c^{2}+\frac{27-15}{4} \leq c^{2}+\frac{27 c-15}{4}<c^{2}+8 c+16=(c+4)^{2}$ because $27 c-15<32 c+64=4(8 c+16)$

Therefore, $c^{2}+\frac{27 c-15}{4} \in\left\{(c+1)^{2} ;(c+2)^{2} ;(c+3)^{2}\right\}$.

By testing each of these cases, we obtain $c=1, c=\frac{31}{11}$ (impossible) or $c=17$.

This leads to the two solutions $(1,1,1)$ and $(37,25,17)$ (which we obtain and verify by substituting the integer values of $c$ into the expressions for $a$ and $b$).
Find all positive integers $a \geqslant b \geqslant c$ such that $a^{2}+3 b, b^{2}+3 c, c^{2}+3 a$ are all perfect squares.