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?

- Thread starter D1Kuta
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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?

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- Jan 29, 2005

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If there were a positive integer solution \(\displaystyle x\) then \(\displaystyle y\) must also be an integer.Prove there are no positive integer solutions to x^2+x+1=y^2 by contradiction.

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

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 can you make up a contradiction proof using this?

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If \(\displaystyle x=2\) then \(\displaystyle x(x+1)\) isHow about x(x+1) must be odd while (y-1)(y+1) must be even? Why is this true?

If \(\displaystyle x=3\) then \(\displaystyle x(x+1)\) is

If \(\displaystyle y=2\) then \(\displaystyle (y-1)(y+1)\) is

If \(\displaystyle y=3\) then \(\displaystyle (y-1)(y+1)\) is

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I did not so any of this. Someone else posted it. I will never admit that I wrote this!If \(\displaystyle x=2\) then \(\displaystyle x(x+1)\) isEVEN not ODD.

If \(\displaystyle x=3\) then \(\displaystyle x(x+1)\) isEVEN not ODD

If \(\displaystyle y=2\) then \(\displaystyle (y-1)(y+1)\) isODD not EVEN.

If \(\displaystyle y=3\) then \(\displaystyle (y-1)(y+1)\) isEVEN notODD.