Question

Show that:
x^2 + x + 1 = y ^2
has no positive integer solutions

Answer 1

If x^2+x+1 takes on an perfect square value for any positive integer x, then there will be a postive integer solution. However, x^2+x+1 = (x+1)^2-x. The difference between two x^2 and (x+1)^2 is 2x+1, so (x+1)^2-x will be between x^2 and (x+1)^2 and thus can't be a perfect square.

Answer 2

Isn't
"The difference between two x^2 and (x+1)^2 is 2x+1"
2x^2 - x^2 + 2x + 1 =
x^2 + 2x + 1

Answer 3

Whoops, I don't know how that "two" got in there - I meant just x^2.

Answer 4

hm, an alternate solution calls for the following assume the equation is true. Then, x<y. By manipulation, we get x^2-y^2 = 1 + y or
(x-y)(x+y) = 1 + y
Since
y>x≥1
Here's where it's unclear for me: we get
x-y > 1 and x + y > 2 + y

Answer 5

Then you would prove that
1*(2+y) becomes a contradiction with the right hand side of the given equation

Answer 6

and also why isn't y considered?