Diophantine equation $2^x-3^y=2021$

Question

$$2^x-3^y=2021$$ where $x,y$ are non-negative integers. I only found $2^{11}-3^3=2021$.

Answer 1

Hmm, three answers with -1 vote. This is interesting.

The fact that we have a solution with $y=3$ suggests that an exclusionary proof must focus on $y\ge 4$ so that the power of $3$ is a multiple of $81$.

So, assume $y\ge 4$ and render

$2^y\equiv 2021\equiv-4\bmod 81.$ (1)

The Euler totient if $81$ is $54$. We have $2^2=4$ and $2^{27}\equiv-1\bmod 81$, so the congruence (1) implies

$x\equiv29\bmod54.$ (2)

This implies that $x$ is two greater than a multiple of $3$ so $2^x\equiv4\bmod 7$. In addition $2021\equiv5\bmod7$, so

$3^y\equiv-1\bmod7,y\equiv3\bmod6.$ (3)

From (3) we find that $y$ has to be a multiple of $3$, so render $y=3z$ from which

$2^x-27^z=2021; x,z\in\mathbb{N}.$ (4)

Now we try modulo ... $109$. Why this number? Experimenting with various other candidates led to the conclusion that the sought contradiction can be realized only if we choose a modulus whose Euler totient is a multiple of $27$, and $109$ is the smallest such prime number.

With $x\equiv29\bmod54$ from (2) and the Euler totient of $109$ being $108$, we have $2^x=\pm2^{29}\equiv\pm23\bmod109$. We also have $2021\equiv59\bmod109$ and thus two possibilities for $27^z$:

$27^z\equiv(23-59)\equiv73\bmod109.$

$27^z\equiv(-23-59)\equiv27\bmod109.$

The first of these results fails, but the second holds when

$z\equiv1\bmod9.$ (5)

Our next prime modulus candidate is $\bmod163$. This has an Euler totient of $162=3×54$, so we have three possible residues for $2^x$ allowed by (2):

$2^x\in\{73,94,159\}\bmod163.$

With $2021\equiv65\bmod163$ we then have:

$27^z\in\{8,29,94\}\bmod163.$

In this case the residues $29$ and $94$ fail whereas $8$ allows $z\equiv23\bmod54$ ($3$ is a primitive root $\bmod163$ so $27^z$ will assume $54$ different residues). But this implies

$z\equiv5\bmod9,$ (6)

whereas above, we found $z\equiv1\bmod9$ (5). The contradiction between (5) and (6) implies that if $y\ge4$, then Eq. (4) cannot hold both $\bmod109$ and $\bmod163$, and without any solution to that equation we can't solve our given equation $2^x-3^y=2021\bmod7$!

We conclude that $2^x-3^y=2021$ has no solutions with $y\ge4$, assuring after trials for smaller $y$ that $(x,y)=(11,3)$ is unique.

Answer 2

According to Pillai, the equation $2^x-3^y=d$ has at most one solution except for $d=-1,5,13$. Since we have a solution for $d=2021$, it is the only solution. For a proof by elementary methods see the reference below (the corollary to Theorem $4$).

Reference: R. Scott, On the Equations $p^x-b^y=c$ and $a^x+b^y=c^z$, J. Number Theory 44 (2) (1993), 153–165.