# Diophantine equation $15^x+8^y=17^z$

How to solve the following equation over the integers?

$$15^x+8^y=17^z$$

I know that only solution is $$(x,y,z) = (2,2,2)$$, but how to prove this?

• $17^z = 15^z + 2^z + P(15, 2, z)=15^x + 2^y+6^y +P(2, 6, y)$ where $p(15, 2, z)$ and $p(2,6,y)$ are polynomials. Since $(15, 2) = 1$ you may assume $15^x = 15^z$ and$2^z = 2^y$and $P(15, 2, z)=6^y+P(6, 2, y)$.That is $x=y=z$. – sirous Oct 29 '17 at 16:58
• Could you please explain in a more detailed way? – Ana Oct 30 '17 at 10:08
• It is binomial expansion of $17^z$ and $8^y$. For example $17^z=(15+2)^z =15^z+2^z+\Sigma C^z_r.15^{z-r}.2^r$.I have denote $\Sigma . . .$as$P(15, 2, z)$ . – sirous Oct 30 '17 at 14:47

By reducing modulo $$7$$ we get $$1+1\equiv 3^z\pmod 7$$, but $$2\equiv 3^2\pmod 7$$ hence $$3^2\equiv 3^z\pmod 7$$ from which $$z\equiv 2\pmod 6$$ because $$3$$ has order $$6$$ in $$(\Bbb Z/7\Bbb Z)^\times$$.

By reducing modulo $$8$$ we get $$(-1)^x\equiv 1\pmod 8$$ from which $$x\equiv 0\pmod 2$$.

By reducing modulo $$15$$ we get $$2^{3y}\equiv 2^z\pmod{15}$$ from which $$3y\equiv z\pmod 4$$ that's $$y\equiv -z\pmod 4$$.

In particular, we have $$x\equiv y\equiv z\equiv 0\pmod 2$$ thus we can write $$x=2\bar x$$, $$y=2\bar y$$ and $$z=2\bar z$$ for positive $$\bar x,\bar y,\bar z$$. Consequently, we get \begin{align} 2^{6\bar y} &=8^y\\ &= 17^z-15^x\\ &=17^{2\bar z}-15^{2\bar x}\\ &=(17^{\bar z}-15^{\bar x})(17^{\bar z}+15^{\bar x}) \end{align} Then there exists $$u,v$$ such that $$u+v=6\bar y$$ and \begin{align} &17^{\bar z}+15^{\bar x}=2^u& &17^{\bar z}-15^{\bar x}=2^v\tag1 \end{align} Note that $$u>v>0$$. Summing we get $$2\cdot 17^{\bar z}=2^u+2^v$$ from which $$17^{\bar z}=2^{v-1}(2^{u-v}+1)$$ from which follows $$v=1$$ because the LHS is odd, consequently, $$u=6\bar y-1$$ and $$17^{\bar z}-1=2^{6\bar y-2}\tag 2$$ By reducing modulo $$15$$ the second equation in $$(1)$$ we get $$2^{\bar z}\equiv 2\pmod{15}$$, from which $$\bar z\equiv 1\pmod 4$$, thus $$\bar z$$ is odd. Hence $$17^{\bar z}-1=2^4\sum_{k=0}^{\bar z-1}17^k\tag 3$$ and comparing with $$(2)$$ gives \begin{align} 2^{6\bar y-6} &=\sum_{k=0}^{\bar z-1}17^k\\ &\equiv\sum_{k=0}^{\bar z-1}1\\ &\equiv\bar z\\ &\equiv 1\pmod{2} \end{align} which implies $$\bar y=\bar z=1$$. This gives $$\bar x=\bar y=\bar z=1$$ hence $$x=y=z=2$$ as unique solution.

• How did you get the first congruence and 2≡3^z(mod7) and how did you convert it to the second congruence z≡2(mod6) – SuperMage1 Nov 2 '17 at 13:47
• Note that $15\equiv 8\equiv 1\pmod 7$. – Fabio Lucchini Nov 2 '17 at 13:51
• Why did you use 17^3 for the congruence, I thought z = 2 mod 6, Also, Why is 15^x = 2^717x (sorry if i'm asking stupid questions) – SuperMage1 Nov 2 '17 at 14:10
• Wrong answer. $2^{2312m+1156}\equiv -1\pmod{17^3}$, $m\ge 0$, $m\in\mathbb Z$. $2312=\frac{\phi(17^3)}{2}$. WolframAlpha says this. – user236182 Nov 2 '17 at 14:20
• @user236182: thank'you for your comment. I edit my answer with a new proof. – Fabio Lucchini Jan 20 '18 at 21:14

Disclaimer: This answer is very similar to Fabio Lucchini's excellent answer, I do not mean to claim his ideas as my own. I was trying to write up an alternative answer, but after running into many dead ends the proof ended up similar to his.

Assuming that $$x$$, $$y$$ and $$z$$ are positive integers, reducing mod $$7$$ and mod $$8$$ yields the congruences $$1^x+1^y\equiv3^z\pmod{7}, \qquad\text{ and }\qquad 7^x+0^y\equiv1^z\pmod{8},$$ which shows that $$z$$ and $$x$$ are even, in that order. Let $$x':=\tfrac{x}{2}$$ and $$z':=\tfrac{z}{2}$$ so that $$8^y=17^{2z'}-15^{2x'}=(17^{z'}-15^{x'})(17^{z'}+15^{x'}).\tag{1}$$

Suppose toward a contradiction that $$x',z'>1$$. Then the two factors on the right hand side are coprime, so $$17^{z'}-15^{x'}=1 \qquad\text{ and }\qquad 17^{z'}+15^{x'}=8^y,$$ and reducing the former mod $$8$$ yields a contradiction. Hence either $$x'=1$$ or $$z'=1$$.

If $$z'=1$$ then $$(1)$$ shows that $$17^{z'}-15^{x'}>0$$, and so $$x'=1$$. So either way have $$x'=1$$, and hence by $$(1)$$ $$17^{z'}-15=2^a \qquad\text{ and }\qquad 17^{z'}+15=2^b,$$ for integers $$b>a\geq0$$ with $$a+b=3y$$. Taking the difference shows that $$2^a(2^{b-a}-1)=2^b-2^a=(17^{z'}+15)-(17^{z'}-15)=30,$$ and hence $$a=2$$ and $$b=5$$, so $$y=2$$. This leaves us with $$15^2+8^2=17^z,$$ with the unique solution $$z=2$$. This proves that $$x=y=z=2$$ is the unique solution.