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.)
If a number n can be expressed as sum of two perfect squares prove that 2^{17^{17^{17^{17}}}}n can also be expressed as sum of two 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.)
Let n be equla to a2+b2
now 217171717 can be written as 2171717(16+1) which can be written as 217171716multiplied by 2171717. Now the former term can be expressed as a perfect square.Similarly if go on doing this we will be left with....
217171716x2171716x21716x216x2x(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