Number of ordered Pairs satisfying $4^m-3^n=1$

Question

Find the Number of ordered Pairs $(m,n)$ of positive integers satisfying $4^m-3^n=1$

Mt try: Trivially $m=n=1$ satisfies

Let $m \gt 1$ $$4^m-3^n=(1+3)^m-3^n=1$$ $\implies$ $$3\binom{m}{1}+3^2\binom{m}{2}+3^3\binom{m}{3}+\cdots+3^m=3^n$$

Now since LHS is not a power of $3$ and RHS is, this is possible only when $m=1$

Hence the only ordered pair is $(1,1)$

is this the right way?

Answer 1

Let $(m,n)$ be a pair with $n>1$ such that $4^m-3^n=1$. Looking at both sides modulo $4$, we see that $n$ must be odd, so of the form $n=2k+1$. Thus we now have $4^m-3\cdot 9^k=1$.

Now looking at the equation modulo $9$, we see that $m$ must be a multiple of $3$, so of the form $m=3l$. Thus we have $64^l-3\cdot 9^k=1$.

However, comparing both sides modulo $7$, we now must have $1-3\cdot 2^k\equiv 1\pmod 7$, or $2^k\equiv 0\pmod 7$. This is clearly impossible, hence $(1,1)$ in fact is the only solution.

PS: I use $n>1$, or, equivalently, $k>0$, to say $3\cdot 9^k\equiv0\pmod9$.

Answer 2

Using a binomial expansion of $4^m=(1+3)^m$ is a perfectly reasonable idea to try, but you can't conclude that the RHS is not a power of $3$ just because it doesn't look like a power of $3$ -- it's conceivable that some choice of $m$ would reel in all the terms to a single power. There might be some way to argue this can't happen, but it's not clear (to me, at least) what form such an argument would take.

An easier proof is available in any event: Look at things mod $8$. We have

$$1+3^n\equiv \begin{cases} 2\mod 8&\text{if $n$ is even}\\ 4\mod 8&\text{if $n$ is odd}\\ \end{cases}$$

whereas $4^m\equiv0$ mod $8$ if $m\ge2$, thus leaving $4^1-3^1=1$ as the only solution.

Remark: I find a lot to like in Kenta S's multi-modulus approach as well; some problems cannot be solved otherwise. It's good to have as large a toolbox as possible, even if you wind up favoring some tools over others.

Answer 3

$4^{m} - 3^{n} = 1$ or $4^m-1=3^n$

$(2^m-1)(2^m+1)=3^n$ therefore $2^m-1$ and $2^m+1$ are perfect powers of 3

Let $2^m+1=3^x$ and $2^m-1=3^y$

$3^x-3^y=2 (x>y)$ therefore $3^y(3^{x-y} -1) = 3^0.2^1$

Hence $ y=0$ and $3^{x-y} -1 =2$ implying $x=1$

Answer 4

The exponent $m$ in the equation $$4^m-1 = 3^n \tag 1$$ must, by Fermat's little and Euler's totient theorem, be of the form $$ m = x \cdot 3^{n-1} \tag 2$$ that means,

• for $n=1 \to m=1$ we have the first solution.

• For $n=2$ we need, that $m=x\cdot 3$ and more over proceed for increasing $n$ as eq (2) indicates

• So for $n=3$ we have already $m=x\cdot 3^2$

A consequence is, that the lhs grows much more than the rhs, and in fact we must rewrite eq 1 for varying $n$ and $m$ depending on $n$ as $$4^{x \cdot 3^{n-1}}-1 = y\cdot 3^n \qquad \text { where } 3 \not | x,y \tag {1a}$$ If $n=2$ we have explicitely $$4^{x \cdot 3}-1 = y\cdot 3^2 \tag {4a}$$ and the smallest nontrivial solution is for $x=1$ so we have

$$ 4^3 -1 = 63 = y \cdot 9 \implies y=7 \\ \implies y \gt 1 \\ \implies \text {eq1 for n=2 impossible} \tag {4b} $$ From eq (1) in the form of (1a) it can, in my view, directly be seen, that increasing $n$ increases $m$ exponentially, and requires $y \gt 1$, but from eq (4a) and eq (2) one can as well start a mathematical induction.

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Notes:

• there is a more-or-less common term "Lifting the exponent" or "LTE"-Lemma which describes the logic of eq (2)

• you might also have a look at my treatize on the same matter in a very general form, see pdf "Cyclic subgroups" on my webspace