# Solve this equation $12^x-5^y=19$ positive integers

Find all $$x,y$$ be positive integers,such $$12^x-5^y=19$$

I found $$(x,y)=(2,3)$$ is solution,maybe have other,so I consider case $$x,y>3$$ and $$\pmod 9$$,since $$12^x\equiv 0\pmod 9,x\ge 2$$

then $$5^y\equiv -1\pmod 9$$,

$$5^1\equiv 5 \pmod 9$$

$$5^2\equiv 7 \pmod 9$$

$$5^3\equiv 8 \pmod 9$$

$$5^4\equiv 4 \pmod 9$$

$$5^5\equiv 2 \pmod 9$$

$$5^6\equiv 1 \mod 9\Rightarrow y\equiv 3 \mod 6\Rightarrow y$$ is odd.

• If you want to eventually prove that there are no other solutions, working mod $9$ will not be enough. You need to work mod $27$ (or mod $8$) because that distinguishes the known solution with $x=2$ from $x\ge 3$. Commented Mar 31, 2022 at 5:12

To show that there are no solutions with $$x \ge 3$$, let's consider the equation modulo $$27$$. When $$x\ge 3$$, the equation becomes $$-5^y \equiv 19 \pmod{27}$$, or $$5^y \equiv 8 \pmod{27}$$. This calculation is intimidating, but rewarding: $$5$$ has multiplicative order $$18$$ modulo $$27$$, and so we are able to conclude $$y \equiv 15 \pmod{18}$$. In particular, $$y \equiv 6 \pmod 9$$.

To make use of this, we look at the factors of $$5^9-1$$: two reasonably-sized ones are $$19$$ and $$31$$. We want factors of $$5^9-1$$ because when $$d \mid 5^9-1$$, knowing that $$y \equiv 6 \pmod 9$$ tells us that $$5^y \equiv 5^6 \pmod d$$.

• Taking the equation mod $$19$$, we get $$12^x - 5^6 \equiv 0 \pmod{19}$$, or $$12^x \equiv 7 \pmod{19}$$. This is true when $$x \equiv 4 \pmod 6$$.
• Taking the equation mod $$31$$, we get $$12^x - 5^6 \equiv 19 \pmod{31}$$, or $$12^x \equiv 20 \pmod{31}$$. This is true when $$x \equiv 2 \pmod {30}$$.

But these two contradict each other, so there are no solutions (with the assumption $$x\ge3$$ we started from).

We could also have combined the last two steps into a calculation modulo $$19\cdot31 = 589$$, where $$12^x - 5^6 \equiv 19 \pmod{589}$$ has no solutions.

Although @MishaLavrov has already provided a solution, I'll show here a pretty mechanical (and general!) method how to detect, that a second solution for $$12^x - 5^y = 19 \tag 1$$ beyond $$(x,y)=(2,3)$$ cannot exist. 1)

We find easily that $$19 = 144 - 125 = 12^2-5^3$$. Thus we have indeed $$12^x - 5^y = 12^2-5^3 \tag 2$$ This can be reformulated into $$12^x-12^2 = 5^y-5^3$$ $$\qquad$$ and then $${12^u-1 \over 5^3 } = {5^v-1 \over 12^2} \tag 3$$ $$\qquad \qquad$$ Here we want, that $$(u=x-2,v=y-3) \gt (0,0)$$.

In the lhs, the numerator to have the primefactor $$5$$ to the power of $$3$$ we must have $$u = 4 \cdot 5^2$$ or a multiple of this, so we could introduce this restriction on $$u$$ $${12^{4 \cdot 5^2 \cdot u_1}-1 \over 5^3 } = {5^v-1 \over 12^2} \tag {4a}$$ In the rhs we can determine a restriction for $$v$$ analoguously: to have the primefactors $$12^2=2^4 \cdot 3^2$$ in the numerator, we must have that $$v=2^2 \cdot 3$$ or a multiple of this, so we introduce this restriction as well: $$\underset{\text{lhs}}{\underbrace{ {12^{4 \cdot 5^2 \cdot u_1}-1 \over 5^3 } }} =\underset{\text{rhs}}{\underbrace{ {5^{2^2 \cdot 3 \cdot v_1}-1 \over 12^2} }} \tag {4b}$$ Now, if $$u_1=1$$ and $$v_1=1$$ have their minimal values, we have in the lhs and the rhs the following primefactor-decompositions: $$\begin{array} {ll} \text{lhs}:& & 13\;\; .11.29.101.1201.1951.19141.22621.60601.73951.\text{} \\ \text{rhs}:& & 13\;\; .7.31.601 \end{array} \tag 5$$ The difference between the primefactorization of the lhs and of the rhs must now be compensated by expanding the exponents $$u_1$$ and $$v_1$$ appropriately to get equality.

But before we start to do this (possibly iteratively) we would look, how an unsolvable contradiction for the equation between lhs and rhs could possibly occur: if the involved adaption of exponents/primefactors includes either, that the lhs get one more primefactor of $$5$$ such that it becomes to be divisible by $$5^4$$ instead of $$5^3$$ - or that the rhs is forced to include exponents in $$v_1$$ such that the (composite) factor $$12$$ occurs to the third power instead of second power (or simply the primefactor $$2$$ to the $$5$$'th power or the primefactor $$3$$ to the third power which would already suffice for an "unsolvability contradiction").

To make it short, we find in the lhs in (4b) (resp. (5)) the primefactor $$p=1201$$. Thus we must expand the exponent in the rhs such that it shall occur as well there. For this, we must expand the exponent to include the "multiplicative order" $$\pmod 5$$ of $$p=1201$$ which is $$o_5=600$$. If we include this value $$o_5$$ into the exponent in the rhs we get the primefactorization

$$\begin{array} {} \Large \text{rhs}&=& \Large {5^{\text{lcm} (2^2 \cdot 3, 600)}-1 \over 12^2} \\ &=& \quad 2 \quad .7.11.13.31.41.61.71.101.151.181.241... \\ &&.251.313.401.521.601.\quad 1201 \quad.1741.1901.\text{} \end{array} \tag 6$$ We have thus indeed adapted the rhs to contain the primefactor $$1201$$.

But, as a sideeffect, we have also one more primefactor $$2$$, and this makes the whole rhs being even.

This expresses contradiction, or condition of impossibility: the lhs can never be even (except both lhs & rhs are zero), so because the primefactor $$p=1201$$ is unavoidedly in the lhs, it must as well be in the rhs, and if it is in the rhs, then as a "collateral effect" the rhs moreover becomes even, which the lhs cannot be.

So this is enough to prove there is no additional solution beyond $$(x,y)=(2,3)$$.

1)The principle of this method has been applied several times already here in MSE at least by Will Jagy and by myself, perhaps later I can add some links.