Hi again,

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

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

Thanks

Pro

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- August 23rd 2011, 11:17 AMprocyonon the subject of a²+b²=2c²
Hi again,

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

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

Thanks

Pro - August 23rd 2011, 11:59 AMSironRe: on the subject of a²+b²=2c²
I don't get your point. What's the intention of the substitutions ?

- August 23rd 2011, 12:20 PMprocyonRe: 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

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)

and you will see that

I am trying to find a similar way to break down

- August 23rd 2011, 12:38 PMprocyonRe: 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

Has the same result since

The trouble is

is not Pythagoras - August 23rd 2011, 02:04 PMmeleseRe: on the subject of a²+b²=2c²
Let be positive integers of opposite parity and ; also, let and be relatively prime.

Then, all solutions of are precisely given by , , , where is any positive integer. Namely, any pair satsifying the above restrictions and any 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 will not show, and still ).

It happens that the question on*(*)*has an answer depending on the above.

From equation*(*)*, we see that are both odd or both even. So, the equation is equivalent to (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)**, , . To find and explicitly we add and subtract and to get , .

**(2)**, , . In a similar way, , .

Example, I choose and apply part**(1)**: Then . Indeed, .(Nod)

Or, if we applied part**(2)**with : and it's true that . - August 23rd 2011, 02:23 PMprocyonRe: on the subject of a²+b²=2c²
oops (Headbang)

Excellent melese, it was so easy in the end.

Many thanks for your help

Pro