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

**chipai** · March 22nd 2009, 05:14 PM

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!!!

**PaulRS** · March 22nd 2009, 05:47 PM

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: $<br /> \left( {2a + 1} \right)^2 + 1= 2c^2 <br />$

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

)

**Soroban** · March 22nd 2009, 07:12 PM

Hello, chipai!

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:

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

The recursive formulas are:

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

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