(a-b)²-c²=2b²

[procyon]

(a-b)²-c²=2b²

\(\displaystyle (a-b)^2-c^2=2b^2\)

\(\displaystyle a>b>0\) and \(\displaystyle a>c>0\) and \(\displaystyle b\not=c\)

From gathered data it appears that the above equation only has +ve integer solutions when b is even.

I've tried re-arranging it in various ways to be able to prove this but can't.

\(\displaystyle c^2=(a-b)^2-2b^2\)

\(\displaystyle c^2=a^2-2ab-b^2\). (2)

From my point of view I can see nothing that implies this property of b.

From (2)

(i) if a even and b even then c will be even
(ii) if a is even and b is odd then c will be odd
(iii) if a is odd and b is even then c will be odd
(iv) if a is odd and b is odd then c will be even

From any results I have obtained, only cases (i) and (iii), where a and c are both odd or both even, occur.

Can anyone see something in the equation that shows this?

The quadratic form doesn't seem to help either

\(\displaystyle b=-a+\sqrt{2a^2-c^2}\)

Many thanks,

Pro

 

[procyon]

The quadratic form doesn't seem to help either

\(\displaystyle b=-a+\sqrt{2a^2-c^2}\)

OK, slight progress I think.

Using the quadratic above, \(\displaystyle 2a^2-c^2\) must be a perfect square.

If a is odd then there is a \(\displaystyle u\ge0\) such that \(\displaystyle a=2u+1\)

Similarly, if c is even, there is a \(\displaystyle v>0\) such that \(\displaystyle c=2v\)

so \(\displaystyle 2a^2-c^2\) can be written

\(\displaystyle 2(2u+1)^2-(2v)^2=8u^2+8u+2-4v^2=2(4u^2+4u-2v^2+1)\)

but \(\displaystyle 4u^2+4u-2v^2+1=2(2u^2+2u-v^2)+1\) which is an odd number.

So we have \(\displaystyle \sqrt{2a^2-c^2}\ \ \ as\ \ \ \sqrt{2*oddNumber}\) which will obviously be irrational since \(\displaystyle \sqrt{2}\) is irrational.

\(\displaystyle \therefore\) there will be no rational roots when a is odd and c is even.

I haven't managed to get the same for a being even and c being odd though :???:

Pro