 Wanda Kane

2022-01-15

Prove that the equation
${2}^{x}-1={y}^{2}$ has no integer solutions where $x>1$. macalpinee3

Expert

$y={2}^{x}-1$ has a root for $x=0$, because its asymptotic at $y=-1$, and is continually increasing. However, ${y}^{2}={2}^{x}-1$ tells a different story. It still has a root at $x=0$, but its parabolic in nature in the horizontal direction.
The reason your function has no roots because of your domain of and at $x=1,y=1$. Hence, what can you conclude about the nature of the function and its position in a cartesian plane after $x>1$. Will it every hit the x-axis? movingsupplyw1

Expert

The point of a mathematical proof is not to be mathematical alenahelenash

Expert

Note that every square number is either . It follows that for $x>2$, we have that ${2}^{x}-1=3\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}4$ and hence, cannot be a square. So if there is a solution it must happen when $x\le 2$. We can manually check that the only remaining potential solution of $x=2$ gives no integer solutions for y.