<s>[INST] Prove that there do not exist integers that satisfy the system 
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}

I thought about using a modular arithmetic argument. The given system is equivalent to \begin{align*}y^2 &= 2x+3+z^2\\z^2 &= 2x+5+w^2\\w^2 &= 2x+7+t^2.\end{align*} Thus $y^2 = 6x+15+t^2$. How can we find a contradiction from this?
 [/INST] My solution is similar, in spirit, to Fimpellizieri's, but organized a little differently ...

Suppose $x,y,z,w,t$ are integers such that
\begin{align}
2x + 3 &= y^2 - z^2 \tag{eq1}\\
2x + 5 &= z^2 - w^2 \tag{eq2}\\
2x + 7 &= w^2 - t^2 \tag{eq3}
\end{align}
\begin{align}
\text{Equation } (1)  &\implies y,z \text{ have opposite parity.}\\
\text{Equation } (2)  &\implies z,w \text{ have opposite parity.}\\
\text{Equation } (3)  &\implies w,t \text{ have opposite parity.}
\end{align}
Thus one of the following cases must hold:


*

*$y,w$ are odd, and $z,t$ are even.

*$y,w$ are even, and $z,t$ are odd.


Consider each case separately.

The key is that an even square must be a multiple of 4, and an odd square must be congruent to 1 (mod 8).

First suppose $y,w$ are odd, and $z,t$ are even.
Then $(\text{eq}2) -(\text{eq}1)$ yields:
\begin{align}
&2z^2 - y^2 - w^2 = 2\\
\implies &2z^2 - y^2 - w^2 \equiv 2 \text{ }(\text{mod }8)\\
\implies &{-2} \equiv 2 \text{ }(\text{mod }8)
\end{align}
contradiction.

Next suppose $y,w$ are even, and $z,t$ are odd.
Then $(\text{eq}3) -(\text{eq}2)$ yields:
\begin{align}
&2w^2 - z^2 - t^2 = 2\\
\implies &2w^2 - z^2 - t^2 \equiv 2 \text{ }(\text{mod }8)\\
\implies &{-2} \equiv 2 \text{ }(\text{mod }8)
\end{align}
contradiction.

It follows that the given system of equations has no integer solutions.
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