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.