Show that if the numbers $\sqrt{a^2+b+c+1}$, $\sqrt{b^2+c+a+1}$ and $\sqrt{c^2+a+b+1 }$ are rational, then $a = b = c$

Question

Let $a$, $b$ and $c$ be nonzero natural numbers. Show that if the numbers $\sqrt{a^2+b+c+1}$, $\sqrt{b^2+c+a+1}$ and $\sqrt{c^2+a+b+1 }$ are rational, then $a = b = c$.

My ideas

For those numebrs to be rational the numbers in the radical must be perfect squares.

I wrote them as some petfect squares.

$a^2+b+c+1=x^2$

$b^2+c+a+1=y^2$

$c^2+b+a+1=z^2$

Atfer summing them up

$a^2+b^2+c^2+2a+2b+2c+3=(a+1)^2+(b+1)^2+(c+1)^2$

Supposing that $a=b=c$ we arrive at the conclusion that we must show that ${a+1}^2=x^2$ and so on.

The form we wrote the numbers as is really close to what we must show. Hope one of can help me! Thank you so much!!

Answer 1

It's clear that the numbers in the radical must be perfect squares，you could choose the largest number of $a,b,c$, WLOG, assume $a \ge b \ge c$.

Notice that $$a^2 \lt a^2+b+c+1 \le (a+1)^2$$ you only need to find when the equality holds.

Answer 2

Hint: Assume, WLOG, $a\geq b\geq c$ and consider the implications of $a^2+b+c+1$ being a perfect square.