on the subject of a²+b²=2c²

**procyon** · August 23rd 2011, 11:17 AM

Hi again,

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

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

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

$and\ \ \ b=2mn$

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

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

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

Thanks

Pro

**Siron** · August 23rd 2011, 11:59 AM

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

**procyon** · August 23rd 2011, 12:20 PM

Hi,

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

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

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)

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

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

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

and you will see that

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

I am trying to find a similar way to break down

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

**procyon** · August 23rd 2011, 12:38 PM

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

$a=2m$

$b=m^2-1$

$c=m^2+1$

Has the same result since

$a^2 = 4m^2$

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

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

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

The trouble is

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

is not Pythagoras

**melese** · August 23rd 2011, 02:04 PM

Originally Posted by procyon

Hi again,

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

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

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

$and\ \ \ b=2mn$

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

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

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

Thanks

Pro

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

Then, all solutions of $a^2+b^2=c^2$ are precisely given by $a=k(m^2-n^2)$ , $b=k(2mn)$ , $c=k(m^2+n^2)$ , where $k$ is any positive integer. Namely, any pair $m,n$ satsifying the above restrictions and any $k>0$ 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 $(15,36,39)$ will not show, and still $15^2+36^2=39^2$ ).

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

From equation (*), we see that $a,b$ are both odd or both even. So, the equation is equivalent to $(\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2$ (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) $(\frac{a+b}{2})=k(m^2-n^2)$ , $(\frac{a-b}{2})=k(2mn)$ , $c=k(m^2+n^2)$ . To find $a$ and $b$ explicitly we add and subtract $(\frac{a+b}{2})$ and $(\frac{a-b}{2})$ to get $a=k(m^2-n^2+2mn)$ , $b=k(m^2-n^2-2mn)$ .

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

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

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

**procyon** · August 23rd 2011, 02:23 PM

oops

Excellent melese, it was so easy in the end.

Many thanks for your help

Pro

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