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

• Mar 22nd 2009, 06:14 PM
chipai
Pythagorean triple in (a,a+1,c)
Pythagorean triple is (a,b,c) satisfies $a^2+b^2=c^2$

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

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

thanks!!!
• Mar 22nd 2009, 06:47 PM
PaulRS
We require: $a^2+(a+1)^2=c^2$

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

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

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

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

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

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

And this is a Pell Type equation ! (Read here, if you have any doubts, ask (Nod) ) :)
• Mar 22nd 2009, 08:12 PM
Soroban
Hello, chipai!

Quote:

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

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

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

The first few are:

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

The recursive formulas are:

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