How to solve the following equation over the integers?


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

  • $\begingroup$ $17^z = 15^z + 2^z + P(15, 2, z)=15^x + 2^y+6^y +P(2, 6, y)$ 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 $x=y=z$. $\endgroup$ – sirous Oct 29 '17 at 16:58
  • $\begingroup$ Could you please explain in a more detailed way? $\endgroup$ – Ana Oct 30 '17 at 10:08
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    $\begingroup$ 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$P(15, 2, z)$ . $\endgroup$ – 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.

  • $\begingroup$ 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) $\endgroup$ – SuperMage1 Nov 2 '17 at 13:47
  • $\begingroup$ Note that $15\equiv 8\equiv 1\pmod 7$. $\endgroup$ – Fabio Lucchini Nov 2 '17 at 13:51
  • $\begingroup$ 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) $\endgroup$ – SuperMage1 Nov 2 '17 at 14:10
  • $\begingroup$ 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. $\endgroup$ – user236182 Nov 2 '17 at 14:20
  • $\begingroup$ @user236182: thank'you for your comment. I edit my answer with a new proof. $\endgroup$ – 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.


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