Find all pairs of integers (x, y) satisfying
1 + x² y = x² + 2xy + 2x + y

Peter

I get

x = -1, y = -1
x = 0, y = 1
x = 1, y = -1,
x = 2, y = -7
x = 3, y = 7

as the only solutions. Does this agree with what you had?

For question 3, I get the horrible expression

(2.4^n-1 + 2.4^n-2 + ... + 2.4 + 3) / (4ⁿ - 2.4^n-1 - 2.4^n-2 - ... - 2.4 - 2)

for p/q, which is either wrong, or must surely be able to be simplified, though I don't immediately see how.

James.

x is an integer solution of the quadratic
x² (1 - y) + x(2y + 2) + (y - 1) = 0

So the discriminant must be a square, i.e.

(2y + 2)² + 4(y - 1)²

is a square.

From there, you can mess around with prime factorisation, and I get the same answer as James.

Interestingly, I did this in a different way:

Rearrange to get 4x = (y - 1)(x² - 2x - 1). But x and x² - 2x - 1 are coprime, so x divides y - 1. So y = ax + 1 for some integer a.

Then 4 = a(x² - 2x - 1), so solve x² - 2x - 1 = b for all b dividing 4, and you're done.

James.

Yep thats what I got. I did it in a slightly different way.

Rearrange to get y = 1 + (4x)/(x² - 2x + 1)

For integer solutions, |4x| must be > = |x² - 2x + 1|.

This only occurs for -2 =< x < = 6.

Just test these values, and the solutions follow.

Peter

Hi,

Michael - how did you work it out using prime factorisation?

Thanks,
Olof.

I can't remember, and I am having difficulty in finding how to make that approach work now. Perhaps I made a mistake first time round - strange because I definitely got James' answer. If I see how to do it I'll write back...

Thanks Michael. I had a go at doing it by prime factorisation but just couldn't see how. It sounds like a very interesting method!

Cheers,
Olof.

I'm pretty sure I made a mistake. It looks kind of promising setting the discriminant to a square, but unfortunately I can't see where to go from here. We know:

2y² + 2 = r²

for some integer r. Now defining a = y + 1 and b = y - 1 we have:

a² + b² = r²

So (a,b,r) are Pythagorean. This means we can write:

a = k(p² - q²
b = 2kpq
c = k(p² + q² )

for some k,p,q

Now a - b = 2, so k(p² - 2pq - q² ) = 2

But y is odd so we know k = +-1. If k = 1 this gives us (p - q)² + 2q² = 2 which is quite interesting as we have generated a new (smaller) solution to a very similar equation. If k = -1 then the sign is reversed.

Anyway, all of this really isn't going to lead to a quick solution - James or Peter's methods are definitely the way to tackle the problem. I'll see if I can find the place I wrote down my original solution, so I can pinpoint the error.