# Dealing with Integers Extension

This post stated an easy Number Theory question regarding the number of solutions of the Diophantine equation $x(x+z)=y^2$.

I decided to make the question a bit more interesting:

Characterize all positive integer solutions to $x(x+z)=y^2$ given that $2y-2x\ge z-1$.

Indeed, the following is also true:

Given that $x,y,z$ are positive integers that satisfy both $x(x+z)=y^2$ and $2y-2x\ge z-1$, then it follows that $2y-2x=z-1$.

My solution is elementary, but well-hidden. I await your solutions.

Note by Daniel Liu
5 years, 8 months ago

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The first solution that comes to my mind:

For Diophantine equation $a^2 = bc$ for $a,b,c$ being positive integers, we can parametrize the solutions as $(a,b,c) = (kmn,km^2,kn^2)$ where $k,m,n$ are positive integers and $\gcd(m,n) = 1$. Essentially, take the GCD of $b,c$, which is $k$; this must divide $a$ as well. For the rest, each of $\frac{b}{k}, \frac{c}{k}$ must be squares; an unmatched divisor of $\frac{b}{k}$ must also exist in $\frac{c}{k}$ (to make a square), but this contradicts that the two are now relatively prime (since we have removed their GCD).

Using this to the equation $x(x+z) = y^2$, we have $x = km^2, x+z = kn^2, y = kmn$. Since $z > 0$, we have $kn^2 > km^2$ or $n > m$.

The condition is:

\begin{aligned} 2y-2x &\ge z-1 \\ 1 &\ge z+2x-2y \\ 1 &\ge x + (x+z) - 2y \\ 1 &\ge km^2 + kn^2 - 2kmn \\ 1 &\ge k(n-m)^2 \end{aligned}

We know that $k \ge 1$ as it's a positive integer. We also know that $n > m$, and since $m,n$ are integers, we have $n-m \ge 1$. Thus equality is achieved; we have $k = 1, n-m = 1$. In other words, $(x,y,z) = (m^2, m(m+1), 2m+1)$. This can be verified to satisfy the equation for all positive integer $m$.

- 5 years, 8 months ago

It is clear from $x(x+z) = y^{2}$ that $x$, $x+z$ and $y$ are in G.P. But since they are distinct, it follows that $x. So, $x$, $y$ and $x+z$ are in G.P. with intial term $a=x$ and with common difference $r=\dfrac{x+z}{y}$. Now, $a+ar^{2} > 2 ar$, I mean that the sum of the extremes is greater than twice the middle term in a G.P. with three terms (Equality does not hold since $r>1$.) So,

$(x+z)+x > 2y$ $z > 2y - 2x$ $z-1 \geq 2y -2x$

But given that $2y-2x \geq z-1$. It holds that $2y - 2x = z- 1$.

- 5 years, 8 months ago

Nice solutions Surya Prakash , Ivan Koswara . Here is mine:

Note that $2y-2x\ge z-1\implies y\ge x+\dfrac{z-1}{2}$

However, we also know trivially that $y^2=x(x+z) <\left(x+\dfrac{z}{2}\right)^2$ so we have the following bounds: $x+\dfrac{z-1}{2}\le y Since $y$ is an integer, it follows that $z$ must be odd and $y=x+\dfrac{z-1}{2}$. Thus we have $\boxed{2y-2x= z-1}$

This also means $z=2t+1$. Solving $x(x+z)=\left(x+\dfrac{z-1}{2}\right)^2$ gives $x=t^2$, and finally $y=t^2+t$ so our parameterization is $\boxed{(x,y,z)=(t^2, t^2+t, 2t+1)}$

- 5 years, 8 months ago