$3^n - 2^m = \pm 41$ is not possible. How to prove it? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:44:42Z http://mathoverflow.net/feeds/question/29926 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29926/3n-2m-pm-41-is-not-possible-how-to-prove-it $3^n - 2^m = \pm 41$ is not possible. How to prove it? Luca 2010-06-29T14:45:38Z 2010-06-29T19:13:14Z <p>$3^n - 2^m = \pm 41$ is not possible for integers $n$ and $m$. How to prove it?</p> http://mathoverflow.net/questions/29926/3n-2m-pm-41-is-not-possible-how-to-prove-it/29928#29928 Answer by Wadim Zudilin for $3^n - 2^m = \pm 41$ is not possible. How to prove it? Wadim Zudilin 2010-06-29T15:15:29Z 2010-06-29T15:15:29Z <p>As a valuable hint for solving the problem, I consider the following extract from my lectures on elementary number theory.</p> <p><strong>Theorem</strong> ($\approx1320$; Levi ben Gerson 1288--1344). The equations $$(1) \quad 3^p-2^q=1$$ and $$(2) \quad 2^p-3^q=1$$ have no solutions in integers $p,q>1$, except the solution $p=2$, $q=3$ to equation (1).</p> <p><em>Proof.</em> (1) If $p=2k+1$, then $$2^q=3^p-1=3\cdot9^k-1\equiv2\pmod4,$$ which is impossible for $q>1$.</p> <p>If $p=2k$, then $2^q=3^p-1=(3^k-1)(3^k+1)$ implying $3^k-1=2^u$ and $3^k+1=2^v$. Since $2^v-2^u=(3^k+1)-(3^k-1)=2$, we have $v=2$ and $u=1$. This corresponds to the unique solution $q=u+v=3$ and $p=2$.</p> <p>(2) If $q\ge1$, then $3^q+1$ is not divisible by~$8$. Indeed, if $q=2k$, then $3^q+1=9^k+1\equiv2\pmod8$; and if $q=2k+1$, then $3^q+1=3\cdot9^k+1\equiv4\pmod8$. Therefore $p\le2$, hence $p=2$. The latter implies $q=1$ which does not correspond to a solution.</p> http://mathoverflow.net/questions/29926/3n-2m-pm-41-is-not-possible-how-to-prove-it/29956#29956 Answer by Max Alekseyev for $3^n - 2^m = \pm 41$ is not possible. How to prove it? Max Alekseyev 2010-06-29T19:03:24Z 2010-06-29T19:13:14Z <p>The congruence $3^n - 2^m \equiv 41\pmod{60}$ has no solutions.</p> <p>The congruence $3^n - 2^m \equiv -41\pmod{72}$ has no solutions.</p>