1. ## Pythagorean triple

prove algebraicly the sum of the reciprocals of 2 consecutive odd numbers always produces the leg of a pythagorem triple

$\displaystyle 1/3+1/5=8/15$
$\displaystyle 8^2+15^2=17^2$

let $\displaystyle 2n-1$ and $\displaystyle 2n+1$ be the two odd natural numbers..

$\displaystyle 1/(2n-1)+1/(2n+1)=(2n-1)+(2n+1)/(2n-1)(2n+1)$
$\displaystyle =4n/4n^2 -4n+1$

thats as fathest as i can get

2. Originally Posted by juanfe_zodiac
prove algebraicly the sum of the reciprocals of 2 consecutive odd numbers always produces the leg of a pythagorem triple

$\displaystyle 1/3+1/5=8/15$
$\displaystyle 8^2+15^2=17^2$

let $\displaystyle 2n-1$ and $\displaystyle 2n+1$ be the two odd natural numbers..

$\displaystyle 1/(2n-1)+1/(2n+1)=(2n-1)+(2n+1)/(2n-1)(2n+1)$
$\displaystyle =4n/{4n^2 \color{red}-4n+1}$

thats as fathest as i can get
The part in red is incorrect. $\displaystyle (2x+1)(2x-1)=4x^2-1$. So you should have $\displaystyle \frac{4n}{4n^2-1}$

From here, I'm sure mathematical induction would be useful here to complete the proof.

3. Originally Posted by Chris L T521
The part in red is incorrect. $\displaystyle (2x+1)(2x-1)=4x^2-1$. So you should have $\displaystyle \frac{4n}{4n^2-1}$

From here, I'm sure mathematical induction would be useful here to complete the proof.
so thats the end of the question???
because i put $\displaystyle n=4$ and i did get $\displaystyle 17^2$

The part in red is incorrect. $\displaystyle (2x+1)(2x-1)=4x^2-1$. So you should have $\displaystyle \frac{4n}{4n^2-1}$