a² + b² = 2c²

[procyon]

a² + b² = 2c²


where

and

and a, b and c are all integers


I tried to break this equation down like this

let


so that



then




but then I realised that

so


this is clearly wrong as c is an integer.


Is there any way of breaking down the original equation?

Many thanks, (btw, if someone could let me know how to properly insert the above images would be a great help also )

Pro

 

[pka]

View attachment 1314
where
View attachment 1313
and
View attachment 1312
and a, b and c are all integers

You can learn to use LaTeX.
[tex]2u^2=\sqrt{2c^2} [/tex] gives \(\displaystyle 2u^2=\sqrt{2c^2}\).

I am not sure exactly what you want done.
Here is one solution: \(\displaystyle a=7,~b=1,~\&~c=5\)

Notice that \(\displaystyle a~\&~b\) are both odd or both even.

 

[procyon]

thanks for the quick reply pka and letting me know about the [tex.] [/tex.] tags

I was basically looking for a similar form to the solution for

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

where choosing any arbitrary m, n gives a correct solution

by setting

\(\displaystyle a = m^2 - n^2\)

\(\displaystyle b = 2mn\)

\(\displaystyle c = m^2 + n^2\)
​

instead of trial and error.

Is there any similar way of approaching \(\displaystyle a^2+b^2=2c^2 ?\)

Thanks again

Pro

 

[Deleted member 4993]

View attachment 1314
where
View attachment 1313
and
View attachment 1312
and a, b and c are all integers


I tried to break this equation down like this

let
View attachment 1315 I think this is the problematic step. c = (2u)^1/2 → c is irrational

Is there any way of breaking down the original equation?

Many thanks, (btw, if someone could let me know how to properly insert the above images would be a great help also )

Pro

.

 

[procyon]

I think this is the problematic step. c = (2u)^1/2 → c is irrational

....as I pointed out at the end of my original post, but thanks anyway

 

[TchrWill]

View attachment 1314
where
View attachment 1313
and
View attachment 1312
and a, b and c are all integers

Just wondering whether you ever discovered any means of deriving other solutions.

Considering a Pell Equation of the form x^2 - Dy^2 = +/-1, (D not a square):

Rearranging your given equation into a Pell Equation format, a^2 - 2c^2 = -b^2, the minimum solution is a = 7 and c = 5 leading to b = 1, or 49 - 2(25) = -1.

Other solutions derive from

....................a = [(p + qsqrt(D))^n + (p - qsqrt(D))^n]/2, n being odd in all cases

...................c = [(p + qsqrt(D))^n - (p - qsqrt(D))^n]/2sqrt(D), n being odd in all cases.

The first two subsequent solutions are a = 99 with c = 70 and a = 1393 with c = 785.

 

[procyon]

Hi,

Yes, here is a way to do it. Like the similar way for pythagorean triples, it returns an infinite number of solutions, but not all solutions.

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

as noted earlier in the thread a and b are both even or both odd, so the above can be written

\(\displaystyle \left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{a-b}{2}\right)^2=c^2\)

where we are back to the pythagorean form of \(\displaystyle x^2+y^2=z^2\)

proof:

\(\displaystyle \left(\dfrac{a+b}{2}\right)^2=\dfrac{a^2+2ab+b^2}{4}\)

\(\displaystyle \left(\dfrac{a-b}{2}\right)^2=\dfrac{a^2-2ab+b^2}{4}\)

so \(\displaystyle \left(\dfrac{a+b}{2}\right)^2+\left(\dfrac{a-b}{2}\right)^2=\dfrac{a^2+b^2}{2}=c^2\)

so \(\displaystyle a^2+b^2=2c^2\)