Solution to $|2^m-3^n|=35$?

Question

As in a previuos question I'm wondering about equations like $|A^m-B^n|=C$, where $A,B,C\in\mathbb N$ are given and solutions $m,n\in\mathbb N$ are wanted. As it seems there are no known general method: either there are solutions to be found or there are non existence proofs to be found (I guess by modular arithmetic methods).

The function $\displaystyle f(m)=\min_{n\in\mathbb N}|A^m-B^n|$ is approximately exponential with some pseudo random minimal values when $f(m)$ happens to be small. Therefore, if $A,B,C$ are "small" the probability of a match for a "big" $m$ is very small. It's interesting that this probability somehow is connected to the existence of counter proofs.

I'm interested in cases where there are no "small" solutions and to see how the proofs are constructed.

Is there a proof that $|2^m-3^n|=35$ doesn't have any solutions $m,n\in\mathbb N$?

I've tested for $m,n<500$ without finding a solution to $|2^m-3^n|=35$.

Answer 1

Modulo $85$, powers of $2$ are only equivalent to elements of $\{1,2,4,8,16,32,64,43\}$, while powers of $3$ are only equivalent to elements of $\{1, 3, 9, 27, 81, 73, 49, 62, 16, 48, 59, 7, 21, 63, 19, 57\}$.

It can simply be bashed out that $2^m-3^n \not\equiv \pm 35\bmod 85$ for any integers $m,n$ (as the set of values of $2^m\pm 35 \bmod 85$ is disjoint with the set of values of $3^n$).

Answer 2

$$ 2^m - 3^n = \pm 35. $$ If both $m,n$ odd, we have $2 x^2 - 3 y^2 = \pm 35 \equiv 0 \pmod 7.$ But the discriminant of the binary form is $24,$ and we find Legendre symbol $$ (24|7) = (6|7) = -1. $$ Which means that we cannot have success modulo $7$ unless both $x,y$ are divisible by $7,$ which does not apply here, $x$ is a power of $2$ and $y$ is a power of $3.$ See proof of Proposition at Prime divisors of $k^2+(k+1)^2$

Similar for one of them odd, give me another minute to type. If both are even, difference of squares factors.

If $m$ odd, $n$ even, $2 x^2 - y^2 = \pm 35.$ Discriminant is $8,$ and $$ (8|5) = (3|5) = -1. $$

If $m$ even, $n$ odd, $ x^2 - 3y^2 = \pm 35.$ Discriminant is $12,$ and $$ (12|5) = (2|5) = -1. $$

If both are even, we have $2^{2t} - 3^{2u} = (2^t + 3^u)(2^t - 3^u)= \pm 35.$ Here we use inequalities, since the possibilities are $2^t + 3^u = 1,5,7,35.$ If $2^t + 3^u = 5,$ perhaps $t=u=1$ and $2^t - 3^u = -1,$ no good; or $t=2, u=0$ and $2^t - 3^u = 3,$ no good. If $2^t + 3^u = 7,$ then $t=2, u=1$ and $2^t - 3^u = 1,$ no good. Finally, two possibilities: if $2^t + 3^u = 35,$ perhaps $t=5, u=1$ and $2^t - 3^u = 29,$ no good; or $t=3, u=3$ and $2^t - 3^u = -19,$ no good.

Answer 3

We find, empirically, $$ 35 = 2^3+3^3 \qquad \qquad \text{and} \qquad (35 = 2^5+3^1)$$ The first option, using $2^m - 3^n=35$ from $|2^m-3^n|$ is easy. The second option, using $2^m-3^n=-35$ is not so easy, and I've not completed that part yet.

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First option:
Equalling the first expression for 35 with our equation with unknowns we write $$ 2^m - 3^n = 2^3+3^3$$. Reordering for equal bases on each side gives $$2^m -2^3 = 3^n +3^3 $$ We see, that $m>3$ is required. So let $m=3+u$ then $$2^3(2^u -1) = 3^n+3^3$$ But now, $\pmod {2^3}$, the residue of the lhs is zero but of the rhs one of $\{4,6\}$. So no solutions is possible.

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Second option:
We use the second equation: $$ 3^n-2^m = 35 = 2^5+3^1 $$ Then we can reorder $$ 3^n - 3^1 = 2^5+2^m $$ By the minimal magnitude of the rhs it follows that $n \ge 4$ and from this follows that $m \ge 5$ because $3^4-3 = 78 > 2^5+2^5$. Then we can factor $$ 3(3^u-1) = 2^5(2^v+1) \qquad \qquad \text{ setting $u=n-1$ and $v=m-5$ }$$ Now to have $(3^u-1)$ containing $32=2^5$ as factor we need that $2^3$ is factor of $u$ but not $2^4$ - otherwise we had too many powers of $2$ here. So we need to have $$ 3(3^{8+16u'}-1) = 2^5(2^v+1) $$

Next, to have in the rhs $3$ to the first power, but not to the second, we must have the exponent being odd and in particular of the form $6v' \pm 1$ so we must have exponents fulfilling the equation

$$ 3(3^{8+16u'}-1) = 2^5(2^{ 6v' \pm 1}+1) $$

But having such a regular exponent on the lhs includes also the primefactor $5$ which occurs always when the exponent is divisible by $4$, and this is always the case in this configuration.
On the other hand, for the same primefactor $5$ to appear as well in the rhs, we must have an even exponent and namely of the form $2^{2+4v"}$

So, for to have in the rhs exactly one factor $3$ the exponent must be odd, but to have the rhs also to contain the factor $5$ the exponent must be even - contradiction.

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Comment: well, this version might look no so nice/smooth as the other one which uses only a single modulus $85$ for both options. But we have here smaller moduli ($8$ in the first option and $64$ and $9$ and $5$ in the second option) and we have some path which leads systematically to the contradiction (if it exists) by some sort of "iterative mutually adapting the exponents" and checking the occuring primefactors beginning with the smallest occuring ones.

Answer 4

We need

$2^m\equiv3^n\pmod7\ \ \ \ (1)$

$\iff2^{m+n}\equiv(-1)^n$

Now as the residues of $2^a\equiv2,4,1\pmod7, n$ must be even $=2r$(say)

$\implies2^{m+2r}\equiv1\implies3|(m+2r)\ \ \ \ (i)$

and

$2^m\equiv3^n\pmod5\ \ \ \ (2)$

$\iff2^{m+n}\equiv6^n\equiv1\iff4|(m+n)\ \ \ \ (ii)$

$\implies4|(m+2r)\implies m$ must be even $m=2s$(say)

$\implies2|(s+r)\ \ \ \ (iii)$ and by $(i),3|(s+r)\ \ \ \ (iv)$

By $(iii),(iv)$ we need $6|(s+r)$

Ex:

If $s\equiv5,r\equiv1\pmod6$

Consequently, $2^m-3^n\equiv2^{10}-3^2$ which is clearly divisible by $35$

We have $m+n\equiv0\pmod{12}, 2^m-3^n=\dfrac{2^{-n}-3^n}{2^{12u}}=\dfrac{1-6^n}{2^{12u+n}}$ for some integer $u$

Now as $n$ is even $6^n\equiv1\pmod{35}$

and as $(35,2^{12u+n})=1,35$ will divide $$\dfrac{1-6^n}{2^{12u+n}}$$