<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] You are asking about four $x$ values. These are consecutive, so they assume the four values $0,1,2,3 \pmod 4.$ Let us give them new names,
$$ x_1 \equiv 1 \pmod 4, \; \;  x_2 \equiv 2 \pmod 4, \; \;  x_3 \equiv 3 \pmod 4, \; \;  x_4 \equiv 0 \pmod 4.  $$
$$ n = x_1^2 + y_1^2 =  x_2^2 + y_2^2 = x_3^2 + y_3^2 = x_4^2 + y_4^2 $$
First we prove that $n$ is odd. Assume $n$ is even.
 $x_1^2 \equiv 1 \pmod 8.$ If $y_1$ is odd, then $n \equiv 2 \pmod 8$ is not divisible by $4.$ However, $x_4^2 \equiv 0 \pmod 8.$ With $y_4$ even, we get $n$ divisible by $4.$ The impossibility about $n$ being divisible by 4 and not divisible by 4  contradicts the assumption that $n$ was even. So, actually, $n$ is odd. 
Again, $x_4^2 \equiv 0 \pmod 8.$ With the necessary odd $y_4,$ we have $y_4^2 \equiv 1 \pmod 8,$ so
$$ n \equiv 1 \pmod 8.  $$ 
However, $x_2^2 \equiv 4 \pmod 8.$ With the necessary odd $y_2,$ we have $y_2^2 \equiv 1 \pmod 8,$ so
$$ n \equiv 5 \pmod 8.  $$ 
The inconsistent values $1,5 \pmod 8$ contradict the assumption that there were four  $ x_1 \equiv 1 \pmod 4, \; \;  x_2 \equiv 2 \pmod 4, \; \;  x_3 \equiv 3 \pmod 4, \; \;  x_4 \equiv 0 \pmod 4.  $
In turn, this contradicts the assumption that we can take the $x$ values $x+1, x+2, x+3, x+4.$ 
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