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.

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TopNewestNice 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: \[x+\dfrac{z-1}{2}\le y <x+\dfrac{z}{2}\] 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 \[\boxed{(x,y,z)=(t^2, t^2+t, 2t+1)}\] – Daniel Liu · 1 year, 4 months ago

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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 \(x<y<x+z\). 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 \(2y - 2x = z- 1\). – Surya Prakash · 1 year, 4 months ago

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– Ivan Koswara · 1 year, 4 months ago

No, it's \(x, y, x+z\) in GP, since we have \(\frac{x}{y} = \frac{y}{x+z}\).Log in to reply

– Julian Poon · 1 year, 4 months ago

yeah, I mistaken.Log in to reply

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:

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

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 \(m\). – Ivan Koswara · 1 year, 4 months ago

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