Discrete maths proof [HELP PLEASE]

[D1Kuta]

Prove there are no positive integer solutions to x^2+x+1=y^2 by contradiction.

All I got was that up to was realizing

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

but I got stuck at this bit, not sure what to do next and how to get the solution... can anyone help please?

 

[pka]

Prove there are no positive integer solutions to x^2+x+1=y^2 by contradiction.

If there were a positive integer solution $\displaystyle x$ then $\displaystyle y$ must also be an integer.
Thus you get
$\displaystyle x^2+x+(1-y^2)=0$
Does that mean that $\displaystyle \sqrt{1-4(1)(1-y^2)}$ has to be a even integer?

 

[Steven G]

Prove there are no positive integer solutions to x^2+x+1=y^2 by contradiction.

All I got was that up to was realizing

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

but I got stuck at this bit, not sure what to do next and how to get the solution... can anyone help please?

How about x(x+1) must be odd while (y-1)(y+1) must be even? Why is this true?

How can you make up a contradiction proof using this?

 

[pka]

How about x(x+1) must be odd while (y-1)(y+1) must be even? Why is this true?

If $\displaystyle x=2$ then $\displaystyle x(x+1)$ is EVEN not ODD .
If $\displaystyle x=3$ then $\displaystyle x(x+1)$ is EVEN not ODD
If $\displaystyle y=2$ then $\displaystyle (y-1)(y+1)$ is ODD not EVEN.
If $\displaystyle y=3$ then $\displaystyle (y-1)(y+1)$ is EVEN not ODD .

 

[Steven G]

If $\displaystyle x=2$ then $\displaystyle x(x+1)$ is EVEN not ODD .
If $\displaystyle x=3$ then $\displaystyle x(x+1)$ is EVEN not ODD
If $\displaystyle y=2$ then $\displaystyle (y-1)(y+1)$ is ODD not EVEN.
If $\displaystyle y=3$ then $\displaystyle (y-1)(y+1)$ is EVEN not ODD .

I did not so any of this. Someone else posted it. I will never admit that I wrote this!