Showing that one of three expressions is a perfect square $a$, $b$, $c$ are natural numbers such that
$a^2 + b^2 + c^2 = 1 + 2abc$
Prove that one of $\frac{a+1}{2}, \frac{b+1}{2}, \frac{c+1}{2} $ is a perfect square.
Since at least one of $a,b,c$ has to be odd, WLOG that $a = 2x-1$, where $x$ is a natural number. Then I tried to split the problem into the cases depending on the parity of $b,c$ but I haven't found anything yet.
 A: If we solve the quadratic in $c$, we obtain
$$
c=ab \pm \sqrt{(a^2-1)(b^2-1)} \tag{1}
$$
We see that $(a^2-1)(b^2-1)$ is a perfect square.
Let $d \gt 0$ be the square-free kernel of $a^2-1$, so that $a^2-1=dA^2$ for some positive integer $A$. Then $b^2-1=dB^2$ for some positive integer $B$.
If $x_0^2-dy_0^2=1$ is the fundamental solution of $x^2-dy^2=1$, then we have two exponents $n$ and $m$ such that
$$
a+A\sqrt{d}=z_0^n, b+B\sqrt{d}=z_0^m \tag{2}
$$
where $z_0=x_0+y_0\sqrt{d}$.
Multiplying or dividing the two relations above, we obtain
$$
ab-dAB +(aB+bA)\sqrt{d} = z_0^{n+m},
ab+dAB +(-aB+bA)\sqrt{d} = z_0^{n-m} \tag{3}
$$
If $n$ is even, say $n=2p$ for some integer $p$, we can write $a+A\sqrt{d}=z_1^2$ where $z_1=z_0^p$. If we write $z_1=x_1+y_1\sqrt{d}$, then $x_1^2-dy_1^2=1$, and $a=x_1^2+dy_1^2=2x_1^2-1$, so $\frac{a+1}{2}=x_1^2$ is a perfect square.
Similarly, if $m$ is even we deduce that $\frac{b+1}{2}$ is a perfect square. Finally, if $n$ and $m$ are both odd, then $n\pm m$ are both even and we deduce from (3) and (1) that $\frac{c+1}{2}$ is a perfect square. This finishes the proof.
