# Discrete maths proof [HELP PLEASE]

#### D1Kuta

##### New member
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

##### Elite Member
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?

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#### Jomo

##### Elite Member
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

##### Elite Member
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 .

#### Jomo

##### Elite Member
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!