Prove that there do not exist integers that satisfy the system

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

This can easily be equated to an extension of the pattern I found for your similar question.

Specifically, any sequence of consecutive odd integers which can be partitioned into three groups such that the sum of the group of smaller numbers is an odd number $2$ greater than the sum of the group of middle numbers, which is in turn $2$ greater than the sum of the group of larger numbers, can be used to derive a solution to this system of equations.

Likewise, any solution to your system of equations can be used to derive such a sequence of consecutive odd integers; the two problems are in fact equivalent.

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Given the proofs provided in other answers that the original system of equations is impossible, we now know that it is also impossible to describe a sequence of positive odd integers which can be divided into three partitions as described above.

(I'd be interested in a direct proof of that fact, but don't have time at the moment to puzzle one out.)

Answer 4

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 $x+1, x+2, x+3, x+4.$