If $a^2+2b+c$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares 
Let $a,b,c$ (with $a,b,c>1$) be postive integers,and  such that $\color{#0a0}{\text{$a^2+2b+c$}}$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares.
  Show that:
  $$a+b+c=276$$

We note that
$$a^2+2b+c\ge (a+1)^2,b^2+2c+a\ge (b+1)^2,c^2+2a+b\ge (c+1)^2$$
 A: Without loss of generality let $a=\min(a,b,c)$. Then there are two cases: $a \leq b \leq c$ and $a \leq c \leq b$.


*

*Suppose $a \leq b \leq c$. Then $c^2 < c^2+2a+b \leq c^2 + 3c < (c+2)^2$, hence $c^2+2a+b=(c+1)^2$ yielding $2a+b=2c+1$. We also have $b^2 < b^2+2c+a = b^2+2a+b-1+a \leq b^2+4b - 1 < (b+2)^2$, hence $b^2+2c+a=(b+1)^2$ and $2c+a=2b+1$. However, since $c \geq b$ and $a \geq 1$ this would imply $a=1$, contradiction.

*Suppose $a \leq c \leq b$. Then $b^2 < b^2+2c+a \leq b^2 + 3b < (b+2)^2$, hence $b^2+2c+a=(b+1)^2$ and $2c+a=2b+1$. This implies $4a+2b = 4a+2c+a-1 \leq 7c - 1 < 8c+8$, so $2a+b < 4c+4$ and $c^2 < c^2+2a+b < (c+2)^2$. It follows that $2a+b=2c+1$. Combining this with $2c+a=2b+1$, we find that $(a,b,c)=(a,3a-2,\frac{5a-3}{2})$. Hence $a$ is odd. Writing $a=2t+1$ with $t \geq 1$ yields $(a,b,c)=(2t+1,6t+1,5t+1)$. Now $b^2+2c+a$ and $c^2+2a+b$ are squares by construction so we only need $a^2+2b+c=4t^2+21t+4$ to be a square. For $t \geq 3$ we have $(2t+4)^2 < 4t^2+21t+4<(2t+6)^2$, so it then follows that $4t^2+21t+4 = (2t+5)^2$ and $t=21$. Hence $t \in \{1,2,21\}$. Since $4t^2+21t+4$ is not a square for $t=1$ and $t=2$, we find $t=21$ and $(a,b,c)=(43,127,106)$ with $a+b+c=276$.
