Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$ 
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}

I was thinking about trying to show by contradiction that no such integers exist. The first equation gives $y^2 = 2x+3+z^2$ while the second gives $z^2 = 2x+5+w^2$. How can we find a contradiction from here?
 A: $-1^2 + 0^2 = 0^2 + 1^2 = 1^2 + 0^2$ if you want a trivial example including negative and zero integers.
So $x = -2$ etc.
A: We need to find (or prove that it does not exist) an odd number $a$ that is the difference of two squares:
$$y^2-z^2=a$$
This is quite easy. We only need to write $a$ as the product of two different numbers: $a=mn$. If $m>n$, say, then the easy possbility is that $m=y+z$ and $n=y-z$. We only need to solve for $y$ and $z$ and we are done. The values for $y$ and $z$ are integers because $m$ and $n$ have the same parity (both are odd).
Now the hard step is that in addition to the former equation we have to find another square $w^2<z^2$ such that $z^2-w^2=a+2$.
In addition, the solution for $z$ in both cases has to be the same, of course.
Then:


*

*$y=(m+n)/2$

*$z=(m-n)/2=(u+v)/2$

*$w=(u-v)/2$

*$mn-uv=-2$


Since $u=m-n-v$ we have
$$mn-uv=mn-(m-n-v)v=v^2-(m-n)v+mn=-2$$
If we see this as a second degree equation on $v$, the discriminant (which must be a square) is
$$(m-n)^2-4mn-8=m^2-6mn+n^2-8=(m-3n)^2-8(n^2+1)$$
Let $s=m-3n$. There exists some $t$ such that
$$s^2-t^2=8(n^2+1)$$
Note that $s$ and $t$ are even.
We can say for example (there are more possibilities) that
$$s+t=2n^2+2$$
$$s-t=4$$
Then we have $s=n^2+3$, $t=n^2-1$. Therefore, $m=n^2+3n+3$. Also
$$v=\frac{n^2+2n+3\pm(n^2-1)}2=\begin{cases}n+2\implies u=n^2+n+1\\n^2+n+1\implies u=n+2\end{cases}$$
The correct solution is what makes $v\le u$, that is $u=n^2+n+1$, $v=n+2$.
Then, for any odd value for $n$ we have a solution (at least). The first of Will Jagy's solutions can be obtained for $n=3$.
Namely:
$$\begin{cases}y=\frac{(n+1)(n+3)}2\\z=\frac{n^2+2n+3}2\\w=\frac{(n+1)(n-1)}2\\x=\frac{y^2-z^2-3}2\end{cases}$$
Example: For $n=13$ we have $m=211$, $u=183$, $v=15$. So $y=112$, $z=99$, $w=84$. And from this,
$$1371^2+112^2=1372^2+99^2=1373^2+84^2=1892185$$
A: The differences between consecutive squares are consecutive odd numbers.  So we have:
$y^2-z^2=$ odd number
$z^2-w^2=$ next odd number
Of course if these were reversed (if the word "next" were moved one equation up) then the solution would be simple; any three consecutive numbers would work.  As it is, it's trickier, but certainly not impossible.
Phrasing the equations as I have, a little thought will reveal that any sequence of consecutive odd numbers which can be partitioned such that the smaller numbers add up to an odd number that is $2$ greater than the sum of the larger numbers of the partition, will provide a solution.
What's more, any solution to the original equation will exhibit the characteristic described in the preceding paragraph; the two questions are equivalent.

Borrowing from another existing answer, let's check this solution against my statement above:
$$31^2 + 12^2 = 32^2 + 9^2 = 33^2 + 4^2$$
The odd numbers which comprise the difference between $4^2$ and $9^2$ are $$(4+5), (5+6),(6+7),(7+8),(8+9)$$ or in other terms: $$9,11,13,15,17$$
The odd numbers which comprise the difference between $9^2$ and $12^2$ are $$19,21,23$$
The first five numbers of course add up to $13\times5=65$, and the latter three numbers add up to $21\times3=63$, which illustrates my earlier conclusion.

And, of course, $65$ and $63$ are the differences between $(x+2)^2$ and $(x+3)^2$, and $(x+1)^2$ and $(x+2)^2$, respectively, which tells us that $x$ will equal half of the smaller of these two odd numbers, after the extra $3$ units are subtracted:  $$\frac {63-3} 2=30$$  (This is because the smaller number, $63$, equals $(x+1)+(x+2)$.)
