Proof:
The samllest square above $a^2$ is $a^2+2a+1$. so we must have for $a^2+b+c$ to be a prefect square
$a^2 + b+ c \ge (a+1)^2$
Or $b+c \ge 2a + 1\cdots(1)$
Similarly $c+a \ge 2b + 1\cdots(2)$
And $a + b \ge 2c + 1\cdots(3)$
Adding above 3 equations we must have $2(a+b+c) >= 2(a + b+ c) + 3$ or $0 \ge 3$ which is contradiction
So above is impossible or No solution exists
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