Question

Find integer triples of x, y and z (x; y; z) that are solutions to the system of equations:
x^2 - y^2 - z^2 = 1; y + z - x = 3

Answer 1

I tried to do traditional ways to solve this, but it is way too hard and I had achieved nothing, apart from finding very high powers of unknown numbers ;)

Answer 2

You have 2 equations but three unknowns, therefore you have one degree of freedom meaning you get to arbitrarily assign a number to any variable... well almost :P Anyways, out of just pure randomness, let \(z=0\) and see whether you can find a solution from that. :)

Answer 3

Solved it, going to post the answer after ~8 hours. It doesn't involve any "take z to be 0" (in fact, z isn't equal to 0 for integer solutions ;) ). Found this MSE question to be very useful: http://math.stackexchange.com/questions/1273844/how-to-solve-3-variable-in-2-equation

Answer 4

Nice, I'm looking forward to see the solution

Answer 5

Oh right, sorry. I didn't read the bit about them having to be integer solutions. :P

Answer 6

@ganeshie8, here's the solution:
First we express x in terms of y and z, from the second equation:
x = y + z - 3

Then we square x and get it to be equal to -> z^2 + y^2 + 2yz - 6z - 6y + 9

After that we just replace x in the first equation with that long sequence of numbers to get:

z^2 + y^2 + 2yz - 6z - 6y + 9 - y^2 - z^2 = 1

As we can see, squares cancel out:

2yz - 6z - 6y + 8 = 0

Then, we find that we can factor this into:
(y - 3)(2z - 6) = 10

10 = 5 * 2 or 10 * 1, so 2z-6 must equal either 1, 2, 5 or 10 (same for y-3).

We notice that 2z - 6 = 2(z - 3), so it's always an even number, meaning that it can be either 2 or 10, leaving only 1 and 5 for y - 3. We apply this to find the solutions:

If 2z - 6 = 10; z = 8, y = 4, x = 9;
If 2z - 6 = 2; z = 4, y = 8, x = 9;

However, we have overlooked the fact that 10 as well can be factored to negative numbers, so we have to check for 2z - 6 = -10 and 2z- 6 = -2:

If 2z - 6 = -10; z = -2, y = 2, x = -3;
If 2z - 6 = -2; z = 2, y = -2, x = -3;

So, all the answers together:
1) x = 9, y = 4, z = 8;
2) x = 9, y = 8, z = 4;
3) x = -3, y = -2, z = 2;
4) x = -3, y = 2, z = -2.

Answer 7

Factorization did the magic! Clever !