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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