prove the proposition Explain why x2y2 3 0 not having any

prove the proposition

Explain why x^2+y^2 - 3 = 0 not having any rational solutions (Exercise 20) implies x^2 + y^2 - 3^k = 0 has no rational solutions for k an odd, positive integer.

Solution

We prove by contradiction

Assume there exist:X,Y so that

X^2+Y^2=3^k has solution for odd positive integer k

k=2n+1,n=0,1,2,....

So,

X^2+Y^2=3^{2n}*3

X^2/3^{2n}+Y^2/3^{2n}=3

(X/3^n)^2+(Y/3^n)^2=3

X, Y are rational. Hence, X/3^n and Y/3^n are rational.

Hence, x^2+y^2=3 has rational solutions:X/3^n,Y/3^n which is a contradiction.

Hence, proved that

x^2+y^2-3^k =0 has no rational solution for k an odd integer