Wanda Kane

Answered

2022-01-15

Prove that the equation

$2}^{x}-1={y}^{2$ has no integer solutions where $x>1$ .

Answer & Explanation

macalpinee3

Expert

2022-01-16Added 29 answers

The reason your function has no roots because of your domain of

movingsupplyw1

Expert

2022-01-17Added 30 answers

The point of a mathematical proof is not to be mathematical

alenahelenash

Expert

2022-01-24Added 366 answers

Note that every square number is either $0\phantom{\rule{0.667em}{0ex}}\mathrm{mod}{\textstyle \phantom{\rule{0.167em}{0ex}}}{\textstyle \phantom{\rule{0.167em}{0ex}}}4\text{}\text{or}\text{}1\phantom{\rule{0.667em}{0ex}}\mathrm{mod}{\textstyle \phantom{\rule{0.167em}{0ex}}}{\textstyle \phantom{\rule{0.167em}{0ex}}}4$ . It follows that for $x>2$ , we have that ${2}^{x}-1=3\phantom{\rule{0.667em}{0ex}}\mathrm{mod}{\textstyle \phantom{\rule{0.167em}{0ex}}}{\textstyle \phantom{\rule{0.167em}{0ex}}}4$ 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.

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