numbers and sum of squares

let n = a² +b² (where a,b are natural nos.)

now 2²² n = 16n = 4² (a² +b²) = [ (4a)² +(4b)² ] (here 4a and 4b are also natural nos. )

= c² +d² (c,d are natural nos.)

Sorry.. I have edited the question...

Let n be equla to a²+b²

now 2^17171717 can be written as 2^171717(16+1) which can be written as 2^17171716multiplied by 2¹⁷¹⁷¹⁷. Now the former term can be expressed as a perfect square.Similarly if go on doing this we will be left with....

2^17171716x2¹⁷¹⁷¹⁶x2¹⁷¹⁶x2¹⁶x2x(a²+b²).

now 2(a²+b²) can be written as [(a+b)²+(a-b)²].

So it is a sum of 2 squares.

Am i right??

It bcz of the simple fact that when n is sum of two perfect squares ,so is 2n. Hence so is 2^{k}n , k is any natural number greater than 1

so i didn`t need to do all these things....

both are nice [1]

nice method! http://www.targetiit.com/smilies/1.gif