on the subject of a²+b²=2c²

Hi again,

I'm sure all the experienced users have come across this

$\displaystyle a^2+b^2=c^2$

$\displaystyle let\ \ \ a=m^2-n^2$

$\displaystyle and\ \ \ b=2mn$

$\displaystyle then\ \ \ c=m^2+n^2$

so that you can choose any integers for m and n to get valid numbers for a, b & c.

Has anyone come across anything similar for

$\displaystyle a^2+b^2=2c^2$

Thanks

Pro

Re: on the subject of a²+b²=2c²

I don't get your point. What's the intention of the substitutions $\displaystyle a=m^2-n^2,...$?

Re: on the subject of a²+b²=2c²

Hi,

Slightly off the point, but if you needed sides lengths for some pythagorean triangles, you could either trial and error your way with

$\displaystyle a^2+b^2=c^2$

or you could choose any m and n as given above to get valid results like this:

let m=7 and n=9 (or any other integers you like)

$\displaystyle a=m^2-n^2=7^2-9^2=-32$

$\displaystyle b=2mn=2*7*9=126$

$\displaystyle c=m^2+n^2=7^2+9^2=130$

and you will see that

$\displaystyle 32^2+126^2=130^2$

I am trying to find a similar way to break down

$\displaystyle a^2+b^2=2c^2$

Re: on the subject of a²+b²=2c²

Another method I just found on wolfram for obtaining Pythagorean triples using only 1 integer to start with by setting n to 1 is

$\displaystyle a=2m$

$\displaystyle b=m^2-1$

$\displaystyle c=m^2+1$

Has the same result since

$\displaystyle a^2 = 4m^2$

$\displaystyle b^2 = m^4-2m^2+1$

$\displaystyle c^2 = m^4+2m^2+1$

$\displaystyle so\ \ \ a^2+b^2=c^2$

The trouble is

$\displaystyle a^2+b^2=2c^2$

is not Pythagoras

Re: on the subject of a²+b²=2c²

Quote:

Originally Posted by

**procyon** Hi again,

I'm sure all the experienced users have come across this

$\displaystyle a^2+b^2=c^2$

$\displaystyle let\ \ \ a=m^2-n^2$

$\displaystyle and\ \ \ b=2mn$

$\displaystyle then\ \ \ c=m^2+n^2$

so that you can choose any integers for m and n to get valid numbers for a, b & c.

Has anyone come across anything similar for

$\displaystyle a^2+b^2=2c^2$

Thanks

Pro

Let $\displaystyle n,m$ be positive integers of opposite parity and $\displaystyle n<m$; also, let $\displaystyle m$ and $\displaystyle n$ be relatively prime.

Then, all solutions of $\displaystyle a^2+b^2=c^2$ are precisely given by $\displaystyle a=k(m^2-n^2)$, $\displaystyle b=k(2mn)$, $\displaystyle c=k(m^2+n^2)$, where $\displaystyle k$ is any positive integer. Namely, any pair $\displaystyle m,n$ satsifying the above restrictions and any $\displaystyle k>0$ will produce a Pythagorean Triple.

Note that while what you wrote gives infinitely many solutions, not *all * solutions are produced this way (Example: the triple $\displaystyle (15,36,39)$ will not show, and still $\displaystyle 15^2+36^2=39^2$).

It happens that the question on *(*)*$\displaystyle a^2+b^2=2c^2$ has an answer depending on the above.

From equation *(*)*, we see that $\displaystyle a,b$ are both odd or both even. So, the equation is equivalent to $\displaystyle (\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2$ (simply verify this), where the fractions are actually integers.

The problem has reduced to another problem that we already know to solve.

There are two possibilities:

**(1)** $\displaystyle (\frac{a+b}{2})=k(m^2-n^2)$, $\displaystyle (\frac{a-b}{2})=k(2mn)$, $\displaystyle c=k(m^2+n^2)$. To find $\displaystyle a$ and $\displaystyle b$ explicitly we add and subtract $\displaystyle (\frac{a+b}{2})$ and $\displaystyle (\frac{a-b}{2})$ to get $\displaystyle a=k(m^2-n^2+2mn)$, $\displaystyle b=k(m^2-n^2-2mn)$.

**(2)** $\displaystyle (\frac{a+b}{2})=k(2mn)$, $\displaystyle (\frac{a-b}{2})=k(m^2-n^2)$, $\displaystyle c=k(m^2+n^2)$. In a similar way, $\displaystyle a=k(m^2-n^2+2mn)$, $\displaystyle b=k(n^2-m^2+2mn)$.

Example, I choose $\displaystyle m=5,n=2,k=1$ and apply part**(1)**: Then $\displaystyle a=5^2-2^2+2\cdot5\cdot2=41,b=5^2-2^2-2\cdot5\cdot2=1,c=5^2+2^2=29$. Indeed, $\displaystyle 41^2+1^2=2(29^2)$.(Nod)

Or, if we applied part**(2)** with $\displaystyle m=7,n=6,k=1$: $\displaystyle a=97,b=71,c=85$ and it's true that $\displaystyle 97^2+71^2=2(85^2)$.

Re: on the subject of a²+b²=2c²

oops (Headbang)

Excellent melese, it was so easy in the end.

Many thanks for your help

Pro