Number theory

[Tomekk]

Hello,

Find all positive integer solutions for following equation.

[mismatchtea]

An inelegant solution...
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x(x+1)=y(y^3+y^2+y+1)
x(x+1)=y(y^2+1)(y+1)

Now, just consider all possibilities:
x = 1

x = y

x = y+1

x = y^2+1

x = y(y^2+1)

x = y(y+1)

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

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

With each of these "x = ..." equations, we know x+1 = y(y+1)(y^2+1)/x. So, we have two equations and two unknowns; if we solve all the equations, then we will have all solutions...
I am sure there is a better solution, but sadly I know very little number theory myself...

[tenniskidperson3]

said better solution

Multiply the equation by 4 and add 1. The left side is is a square, so the right side, must be also.

Note that

.

By the first equality, if is a positive integer, then , because is always going to be greater than 0.

By the second equality, if is a positive integer greater than 2, then because . So the only possible case is if or . But if , then what needs to be a square is in fact 17. But when , we need to be a square, and it is, 121. This is so .

Thus the only solution in positive integers is .