Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$

Question

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?

Answer 1

$$ 25 = 5^2 = 3^2 + 4^2 = 4^2 + 3^2 = 5^2 + 0^2$$ $$ 1105 = 5 \cdot 13 \cdot 17 = 31^2 + 12^2 = 32^2 + 9^2 = 33^2 + 4^2$$ $$ 12025 = 5^2 \cdot 13 \cdot 37 = 107^2 + 24^2 = 108^2 + 19^2 = 109^2 + 12^2$$ $$ 66625 = 5^3 \cdot 13 \cdot 41 = 255^2 + 40^2 = 256^2 + 33^2 = 257^2 + 24^2$$ $$ 252601 = 41 \cdot 61 \cdot 101 = 499^2 + 60^2 = 500^2 + 51^2 = 501^2 + 40^2$$ $$ 292825 = 5^2 \cdot 13 \cdot 17 \cdot 53 = 539^2 + 48^2 = 540^2 + 35^2 = 541^2 + 12^2$$ $$ 751825 = 5^2 \cdot 17 \cdot 29 \cdot 61 = 863^2 + 84^2 = 864^2 + 73^2 = 865^2 + 60^2$$

INSERT FRIDAY: in order to show that these examples are always $n \equiv 1 \pmod 8,$ it suffices to consider everything $\pmod{16}.$ Odd squares are $1,9 \pmod {16}.$ Indeed, if $t \equiv \pm 1 \pmod 8$ then $t^2 \equiv 1 \pmod{16}.$ if $s \equiv \pm 3 \pmod 8$ then $s^2 \equiv 9 \pmod{16}.$ The restriction comes in even squares, which are $0, 4 \pmod {16}.$ That is most of the argument, that $12$ is not a square $\pmod {16}.$ Call it a proposition, it is not possible to have $$ n = x_1^2 + y_1^2 = x_2^2 + y_2^2 =x_3^2 + y_3^2 $$ with $$ x_1 \equiv \pm 1 \pmod 8, \; \; \; x_2 \equiv 2 \pmod 4, \; \; \; x_3 \equiv \pm 3 \pmod 8. $$

Except for 292825 there seems to be a pattern, probably one variable polynomials.

INSERT: The pattern is easier to see if we translate the $n$ below as $n=m-1,$ or $m=n+1:$ $$ \color{magenta}{ w = 16 m^6 + 4 m^4 + 4 m^2 + 1} $$ $$ \color{red}{w = \left( 4 m^3 - 1 \right)^2 + \left(2 m^2 + 2m \right)^2} $$ $$ \color{red}{w = \left( 4 m^3 \right)^2 + \left(2 m^2 + 1 \right)^2 }$$ $$ \color{red}{ w = \left( 4 m^3 + 1 \right)^2 + \left(2 m^2 - 2m \right)^2 }$$ Yep, $$ w = 16n^6 + 96n^5 + 244 n^4 + 336 n^3 + 268 n^2 + 120 n + 25 $$ $$ w = \left( 4 n^3 + 12 n^2 + 12 n + 3 \right)^2 + \left( 2n^2 + 6n + 4 \right)^2 $$ $$ w = \left( 4 n^3 + 12 n^2 + 12 n + 4 \right)^2 + \left( 2 n^2 + 4n +3 \right)^2 $$ $$ w = \left( 4 n^3 + 12 n^2 + 12 n + 5 \right)^2 + \left( 2 n^2 + 2n \right)^2 $$

Check algebra:

parisize = 4000000, primelimit = 500509
? u = ( 4 * n^3 + 12 * n^2 + 12 * n + 3)^2  + ( 2 * n^2 + 6 * n + 4)^2
%1 = 16*n^6 + 96*n^5 + 244*n^4 + 336*n^3 + 268*n^2 + 120*n + 25
? v = ( 4 * n^3 + 12 * n^2 + 12 * n + 4)^2  + ( 2 * n^2 + 4 * n + 3)^2
%2 = 16*n^6 + 96*n^5 + 244*n^4 + 336*n^3 + 268*n^2 + 120*n + 25
? w = ( 4 * n^3 + 12 * n^2 + 12 * n + 5)^2  + ( 2 * n^2 + 2 * n + 0)^2
%3 = 16*n^6 + 96*n^5 + 244*n^4 + 336*n^3 + 268*n^2 + 120*n + 25
? u - v
%4 = 0
? v - w
%5 = 0
? w - u
%6 = 0
? 

Answer 2

$-1^2 + 0^2 = 0^2 + 1^2 = 1^2 + 0^2$ if you want a trivial example including negative and zero integers.

So $x = -2$ etc.

Answer 3

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

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, $$1371^2+112^2=1372^2+99^2=1373^2+84^2=1892185$$

Answer 4

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.

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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.

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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 $(x+1)+(x+2)$.)