# Solving the Diophantine Equation $2 \cdot 5^n = 3^{2m} + 1$ over $\mathbb{Z}^+$.

Prove that $(m, n) = (1, 1)$ is the only solution for the Diophantine Equation $$2 \cdot 5^n = 3^{2m} + 1$$ where $(m, n) \in (\mathbb{Z}^+)^2$.

I've managed to prove that both $m$ and $n$ are odd seeing $\bmod 3\text{ and } 10$ respectively. Also, $\forall n \ge 1$, $10$ divides the LHS. I am not able to proceed from here. Any help would be appreciated.

• I'm just being a weenie, but an exponential Diophantine equation is not, technically, a Diophantine equation. – Thomas Andrews Aug 16 '16 at 15:44
• You can also see that $m$ must be odd if you expand $(10-1)^m+1$ – Ovi Aug 16 '16 at 15:55
• @Ovi Simply notice that $9^m+1\equiv (-1)^m+1\equiv 0\pmod{5}$. – user236182 Aug 16 '16 at 15:55
• @user236182 I don't really know any number theory, I just wanted to share the idea of expanding $(10-1)^m+1$. I haven't been able to use it to get a solution but I thought maybe someone else can. – Ovi Aug 16 '16 at 15:57

Here's a non-elementary solution that uses the Zsigmondy's theorem.

If $m\ge 2$, then $9^m+1$ has a prime factor that does not divide $9+1=2\cdot 5$, contradiction. Therefore $m=1$ and so $n=1$.

Another solution can use the Lifting The Exponent lemma. mod $5$ gives that $m$ is odd. Also $5$ is prime, $5\mid 9+1$ and $5\nmid 9$ and $5\nmid 1$.

$$\upsilon_5\left(9^m+1\right)=\upsilon_5(9+1)+\upsilon_5(m)$$

$$\upsilon_5(m)=n-1$$

$$\implies 5^{n-1}\mid m\implies m\ge 5^{n-1}$$

$$2\cdot 5^n=9^m+1\ge 9^{5^{n-1}}+1$$

is wrong if $n\ge 2$, therefore $n=1$, so $m=1$.

Here is an elementary solution. $n = 1$ gives the solution above. Now, suppose $n > 1$. Thus, $9^m + 1 \equiv 0 \pmod {25}$. Looking at all the possible cases, we have $m = 10k + 5$ for integer $k$. We can check that there are no solutions for $m = 2, \ldots, 5$. Thus, $k > 0$. Theefore,

$$2 \cdot 5^n = 3^{2(10k + 5)} + 1 = 3^{10} \cdot 3^{20k} + 1.$$

Now $1181$ divides $3^{20} - 1$. Thus, $3^{20} \equiv 1 \pmod {1181}$ and $3^{10} \equiv - 1 \pmod{1181}$. Now taking the equation $\pmod{1181}$ gives

$$2 \cdot 5^n \equiv -1 \cdot 1 + 1 \equiv 0 \pmod {1181}$$ which has no solutions. Thus $(m,n) = (1,1)$ is the only solution.

• Why 1181? ${}{}{}$ – TheRandomGuy Aug 16 '16 at 15:45
• @Dhruv Because it gets rid of the $3^{20k}$ because $1181$ divides $3^{20} - 1.$ – Sandeep Silwal Aug 16 '16 at 15:46
• I mean how did you find 1181? Did you use a technique or brute force? – TheRandomGuy Aug 16 '16 at 15:49
• @Dhruv: I used wolfram alpha to find that prime factor. – Sandeep Silwal Aug 16 '16 at 15:51

mod $3$ gives $n=2k+1$ is odd.

I'll solve a stronger equation: $x^2-10y^2=-1$, where $x=3^m$, $y$ is any integer.

Here's a long solution that uses Pell equations.

As I said in my answer here (where I also mention where you can see a full solution of a very similar, analogous problem -- solving $5^n=6m^2+1$ in integers), Pell equation $x^2-Dy^2=-1$ has the solutions

$$x_{v}=\pm\frac{(x_1+y_1\sqrt{D})^{2v-1}+(x_1-y_1\sqrt{D})^{2v-1}}{2}$$

$$y_{v}=\pm\frac{(x_1+y_1\sqrt{D})^{2v-1}-(x_1-y_1\sqrt{D})^{2v-1}}{2\sqrt{D}}$$

$v=1,2,\ldots$ Using $v$ instead of $n$ is a good decision here because $n$ already appears in some Diophantine equations.

$(x_1,y_1)=(\pm 3,\pm 1)$.

$$x_{v}=\pm\frac{(3+\sqrt{10})^{2v-1}+(3-\sqrt{10})^{2v-1}}{2}$$

$$y_{v}=\pm\frac{(3+\sqrt{10})^{2v-1}-(3-\sqrt{10})^{2v-1}}{2\sqrt{10}}$$

$v=1,2,\ldots$

$$2\cdot 3^m=(3+\sqrt{10})^{2v-1}+(3-\sqrt{10})^{2v-1}$$

If $m=1$, then we have a solution $(n,m)=(1,1)$. Let $m\ge 2$. Then notice that $2v-1$ is odd, remember the binomial theorem:

$$(3+\sqrt{10})^{2v-1}+(3-\sqrt{10})^{2v-1}$$

$$\equiv 6(2v-1)(\sqrt{10})^{2v-2}$$

$$\equiv 6(2v-1)10^{v-1}\equiv 0\pmod{9}$$

$$\iff 2v-1\equiv 0\pmod{3}$$

$(2v-1)/3$ is odd, therefore $$2\cdot 3^2\cdot 13=(3+\sqrt{10})^3+(3-\sqrt{10})^3\mid$$

$$\mid (3+\sqrt{10})^{2v-1}+(3-\sqrt{10})^{2v-1}=2\cdot 3^m,$$

contradiction. How do I know that $$b_3:=(3+\sqrt{10})^3+(3-\sqrt{10})^3\mid$$

$$\mid (3+\sqrt{10})^{2v-1}+(3-\sqrt{10})^{2v-1}=:b_{2v-1}?$$

((($:=$ denotes "equal by definition")))

Well, the solution of $5^n=6m^2+1$ which I gave in the beginning of this answer uses this

and also see this link from AoPS. I won't be self-contained here.

v_Enhance said that, in this case: "$b_{2v-1}/b_3$ is both rational and an algebraic integer."

Kent Merryfield has a longer answer, which you should see for yourself.