# Thread: Pythagorean triple in (a,a+1,c)

1. ## Pythagorean triple in (a,a+1,c)

Pythagorean triple is (a,b,c) satisfies $\displaystyle a^2+b^2=c^2$

Find all Pythagorean triple in (a,a+1,c)

the first two are (3,4,5) and (20,21,29)

thanks!!!

2. We require: $\displaystyle a^2+(a+1)^2=c^2$

Expand: $\displaystyle 2a^2+2a+1=c^2$

Now multiply by 2: $\displaystyle 4a^2+4a+2=2c^2$

So: $\displaystyle 4a^2+4a+1+1=2c^2$ and note that: $\displaystyle 4a^2+4a+1=(2a+1)^2$

We have: $\displaystyle \left( {2a + 1} \right)^2 + 1= 2c^2$

Set $\displaystyle x=2a+1$ and $\displaystyle y=c$

So our condition turns into: $\displaystyle x^2+1=2y^2$

And this is a Pell Type equation ! (Read here, if you have any doubts, ask )

3. Hello, chipai!

Pythagorean triple is $\displaystyle (a,b,c)$ satisfies $\displaystyle a^2+b^2\:=\:c^2$

Find all Pythagorean triple of the form: $\displaystyle (a,a+1,c)$

The first two are $\displaystyle (3,4,5)$ and $\displaystyle (20,21,29)$.
The first few are:

. . $\displaystyle \begin{array}{ccc} a & b & c \\ \hline 3 & 4 & 5 \\ 20 & 21& 29 \\ 119& 120& 169 \\ 696 & 697 & 985 \\ 4059 & 4060 & 5741 \end{array}$

The recursive formulas are:

. . $\displaystyle \begin{array}{ccc}a_n &=& 6a_{n-1} - a_{n-2} + 2 \\ \\[-3mm] c_n &=& 6c_{n-1} - c_{n-2}\qquad \end{array}$