MathLabs

Problem 4

Determine all pairs (x,y)(x,y) of integers such that 1+2x+22x+1=y21+2^{x}+2^{2x+1}=y^{2}.
Step 3 of 7: Factor for positive xx
(y−1)(y+1)=2x(1+2x+1)(y-1)(y+1)=2^x(1+2^{x+1})
Detailed analysis

Now let x>0x>0. The right side of the original equation is odd, so yy is odd. Rearranging gives (y−1)(y+1)=2x(1+2x+1)(y-1)(y+1)=2^x(1+2^{x+1}). Both factors on the left are even and exactly one is divisible by 44.