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

1. Jan 4, 2006

### 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.

-Steve

2. Jan 4, 2006

### 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?

3. Jan 17, 2006

### 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.

Last edited: Jan 18, 2006