Thread: (x+1)! + (y+1)! = x^2 x y^2

1. (x+1)! + (y+1)! = x^2 x y^2

Hello, any assistance on the problem would be much appreciated. I have a vague idea but cant quite remember what to do.

(x+1)! + (y+1)! = x^2 x y^2, how many pairs such that x and y are a pair of positive, non-negative integers?

Any help would be great, my maths tutor set this and i honestly cant quite remember what to do in this case.
Cheers

2. Re: (x+1)! + (y+1)! = x^2 x y^2

Originally Posted by pikachu26134
Hello, any assistance on the problem would be much appreciated. I have a vague idea but cant quite remember what to do.

(x+1)! + (y+1)! = x^2 x y^2, how many pairs such that x and y are a pair of positive, non-negative integers?

Any help would be great, my maths tutor set this and i honestly cant quite remember what to do in this case.
Cheers
$(x+1)! > x^4$ when $x>4$
$(y+1)! > y^4$ when $y>4$

$(x+1)! + (y+1)! > x^4 + y^4 \geq 2x^2y^2$

4. Re: (x+1)! + (y+1)! = x^2 x y^2

Originally Posted by pikachu26134
ASZ has ruled out almost all the cases for you. The only remaining cases are when $x \leq 4, y\leq 4$. Substitute the remaining values and try to find the answer.

5. Re: (x+1)! + (y+1)! = x^2 x y^2

M=4, N=3
and M=3, N=4
As in 2 pairs such that that equation is true.

6. Re: (x+1)! + (y+1)! = x^2 x y^2

Originally Posted by pikachu26134
M=4, N=3
and M=3, N=4
As in 2 pairs such that that equation is true.

Yes! But instead of using M and N use x and y, or you can write it like that: (3,4) and (4,3) (due to the symmetric equation) .