Dealing with Integers Extension

This post stated an easy Number Theory question regarding the number of solutions of the Diophantine equation x(x+z)=y2x(x+z)=y^2.

I decided to make the question a bit more interesting:

Characterize all positive integer solutions to x(x+z)=y2x(x+z)=y^2 given that 2y2xz12y-2x\ge z-1.

Indeed, the following is also true:

Given that x,y,zx,y,z are positive integers that satisfy both x(x+z)=y2x(x+z)=y^2 and 2y2xz12y-2x\ge z-1, then it follows that 2y2x=z12y-2x=z-1.

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

Note by Daniel Liu
5 years, 11 months ago

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The first solution that comes to my mind:

For Diophantine equation a2=bca^2 = bc for a,b,ca,b,c being positive integers, we can parametrize the solutions as (a,b,c)=(kmn,km2,kn2)(a,b,c) = (kmn,km^2,kn^2) where k,m,nk,m,n are positive integers and gcd(m,n)=1\gcd(m,n) = 1. Essentially, take the GCD of b,cb,c, which is kk; this must divide aa as well. For the rest, each of bk,ck\frac{b}{k}, \frac{c}{k} must be squares; an unmatched divisor of bk\frac{b}{k} must also exist in ck\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)=y2x(x+z) = y^2, we have x=km2,x+z=kn2,y=kmnx = km^2, x+z = kn^2, y = kmn. Since z>0z > 0, we have kn2>km2kn^2 > km^2 or n>mn > m.

The condition is:

2y2xz11z+2x2y1x+(x+z)2y1km2+kn22kmn1k(nm)2\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 k1k \ge 1 as it's a positive integer. We also know that n>mn > m, and since m,nm,n are integers, we have nm1n-m \ge 1. Thus equality is achieved; we have k=1,nm=1k = 1, n-m = 1. In other words, (x,y,z)=(m2,m(m+1),2m+1)(x,y,z) = (m^2, m(m+1), 2m+1). This can be verified to satisfy the equation for all positive integer mm.

Ivan Koswara - 5 years, 11 months ago

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It is clear from x(x+z)=y2x(x+z) = y^{2} that xx, x+zx+z and yy are in G.P. But since they are distinct, it follows that x<y<x+zx<y<x+z. So, xx, yy and x+zx+z are in G.P. with intial term a=xa=x and with common difference r=x+zyr=\dfrac{x+z}{y}. Now, a+ar2>2ara+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>1r>1.) So,

(x+z)+x>2y(x+z)+x > 2y z>2y2xz > 2y - 2x z12y2xz-1 \geq 2y -2x

But given that 2y2xz12y-2x \geq z-1. It holds that 2y2x=z12y - 2x = z- 1.

Surya Prakash - 5 years, 11 months ago

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

Note that 2y2xz1    yx+z122y-2x\ge z-1\implies y\ge x+\dfrac{z-1}{2}

However, we also know trivially that y2=x(x+z)<(x+z2)2y^2=x(x+z) <\left(x+\dfrac{z}{2}\right)^2 so we have the following bounds: x+z12y<x+z2x+\dfrac{z-1}{2}\le y <x+\dfrac{z}{2} Since yy is an integer, it follows that zz must be odd and y=x+z12y=x+\dfrac{z-1}{2}. Thus we have 2y2x=z1\boxed{2y-2x= z-1}

This also means z=2t+1z=2t+1. Solving x(x+z)=(x+z12)2x(x+z)=\left(x+\dfrac{z-1}{2}\right)^2 gives x=t2x=t^2, and finally y=t2+ty=t^2+t so our parameterization is (x,y,z)=(t2,t2+t,2t+1)\boxed{(x,y,z)=(t^2, t^2+t, 2t+1)}

Daniel Liu - 5 years, 11 months ago

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