<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] Let's look at the equations modulo $8$. Notice that the only squares modulo $8$ are $0,1$ and $4$.
Case $(1)$: $x \equiv 0,4 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+3\\
z^2&\equiv w^2+5\\
w^2 &\equiv t^2+7\end{align*}
The first equation implies $y^2\equiv 4$ and $z^2\equiv 1$. This implies $w^2 \equiv 4$ via the second equation. But then the third equation implies becomes $t^2 \equiv 5$, which cannot be.
Case $(2)$: $x \equiv 2,6 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+7\\
z^2&\equiv w^2+1\\\
w^2 &\equiv t^2+3\end{align*}
The first equation implies $y^2\equiv 0$ and $z^2\equiv 1$. This implies $w^2 \equiv 0$ via the second equation. But then the third equation implies becomes $t^2 \equiv 5$, which cannot be.
Case $(3)$: $x \equiv 1,5 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+5\\
z^2&\equiv w^2-1\\
w^2 &\equiv t^2+1\end{align*}
The first equation implies $y^2\equiv 1$ and $z^2 \equiv 4$, but then the second equation cannot be satisfied.
Case $(4)$: $x \equiv 3,7 \pmod 8$
Then we have the system
\begin{align*}
y^2&\equiv z^2+1\\
z^2&\equiv w^2+3\\
w^2 &\equiv t^2+5\end{align*}
The first equation implies $y^2\equiv 1$ and $z^2\equiv 0$, but then the second cannot be satisfied, and this concludes our proof.
</s>