Hey, I need a bit of help with a proof, please

[Morbid Steve]

Here's the item needing to be proofed (this is not homework, but I'm very interested in it).

Show that for each odd prime number y, there is exactly one positive integer x such that x(x+y) is a perfect square.


Thanks for any help/leads, etc..

-Steve

 

[shmoe]

If x(x+y) is a perfect square, show that x and x+p are relatively prime. What can you then say about x?

 

[maverick6664]

hmm...this looks stalled, so I'll write the whole proof
Since x(x+y) is a perfect square, x must be a perfect square.

Proof: x(x+y) can be of a^2b^2 form or a^2 b^2 c^2 form. If x isn't a perfect squre, x can be of one of forms: x=a or x=a^2 b, and then x+y = a b^2 or x+y=b or x+y=b c^2 respectively, but in any case, x and x+y has a common measure. This contradicts the fact y is a prime number.

Now we can write x = a^2 and so x+y=d^2. Then y =(x+y) - x = d^2 - a^2 = (d+a)(d-a). Now that y is a prime number, d-a must be 1 (notice y cannnot be factorized)! So d = a+1 and y = d+a = 2a+1. So x = a^2 = ((y-1)/2)^2 and this is the only possible value of x.

This is just my solution, so there may be better solutions.