Show that there are no positive integer solutions to $x^2 + x + 1 = y^2$ I'm trying to prove that
$$x^2 + x + 1 = y^2$$
has no integer solution. So far I've tried proof by contradiction, but all of that seems to rely on me being able to factor this expression into some neat form where the solution is obvious. I've not been able to do so.
 A: Start by multiplying both sides by $4$.
$$4y^2=4x^2+4x+4$$
$$(2y)^2=(2x+1)^2+3$$
$$(2y)^2-(2x+1)^2=3$$
The only $2$ squares whose difference is $3$ are $4$ and $1$.  So $2x+1=\pm1$, which has no positive solution. 
A: Hint:
If $n$ is square then $n\equiv 1,4,0 (\mod 8) $.
In your case $y$ is an odd number then $$y^2\equiv 1(\mod 8)$$ and 
$$
x^2<y^2<x^2+2x+1
$$
Now it 's easy to conclude. 
A: You could try completing the square to obtain $(x+\frac{1}{2})^2 + 1 - \frac{1}{4} = y^2 $. So if there exist integer solutions to this equation, then $(x+\frac{1}{2})^2 + 1 - \frac{1}{4}$ is a perfect square. Go from here to reach a contradiction.
A: your statement can be  written as 
for all x,y in Z^+ such that (x^2+x+1 is not equal to y^2) 
Negation of this statement should be 
Assume there exists x,y in Z^+ (x^2+x+1=y^2)
Three situations can arise 
1. x=y 
then factorize above equation as (y-x)(y+x)=x+1 
since x=y therefore 0=x+1 => x=-1 which contradicts our assumption.
2. x>y 
then factorize above equation as (y-x)(y+x)=x+1
since x>y therefore l.h.s = negative while R.h.s is positive which contradicts our assumption. 
3. xx+1 we  can also say that (y-x)(y+x)>x+1 which also contradicts our third condition. So our proposition is true.
A: 
$x^2 + x + 1 = y^2$ has no positive integer solutions.

Assume that $x$ and $y$ are positive:  then
$$x^2 - y^2 = -(x+1)$$ 
Therefore,
$$(x-y)(x+y) = -(x+1)$$ ($x$ is positive so the right hand side is positive)


*

*this implies $x >  y$ 
$x= m + y$ ((m some positive integer))
$(y + m)^2 + y + m = y^2$ ((contradiction))  
then $x < y$ which implies that there is contradiction, 
so our statement is true. 
