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The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions: \begin{eqnarray*} (x_1,y_1,z_1)&=&(3,0,1),\\ (x_2,y_2,z_2)&=&(1,1,1)\\ (x_3,y_3,z_3)&=&(5,2,2).\\ \end{eqnarray*} There are no other natural solutions for $z≤10000$.

It seems that the equation $2^{x}+7^{y}=9^{z}$ has no other solutions in natural numbers. How can this be proven?

Eric Wofsey
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CatMario
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5 Answers5

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Hint 1: If $x>1$ then reducing mod $4$ shows that $y$ is even, say $y=2v$, and so $$2^x=(3^z+7^v)(3^z-7^v).$$ Hint 2: It follows that $3^z-2^{x-2}=1$.

Servaes
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    Modulo $80$, we get $x=4k+5$, $y=4l+2$, $z=2m$. It is important that $y$ is even. It doesn't matter what $z$ is, since $9$ is already a square. I also considered your idea, but did not go far. – QLimbo Nov 03 '21 at 21:11
  • @Cornifer I have added another hint. – Servaes Nov 03 '21 at 22:31
  • Why cannot there be an infinite number of solutions w $x=1$? – Mike Nov 03 '21 at 23:30
  • @John Omielan why cant $x=1$? – Mike Nov 04 '21 at 01:00
  • Hi @John Omielan, I knew that, what I meant was, does anyone have a proof that there are not an infinite number of sol'ns for $x=1$. I cited you specifically because you are great at figuring out the challenging number-theory problems on here :) – Mike Nov 04 '21 at 03:08
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    @Mike Sorry about misunderstanding your earlier comment. Also, thanks for the compliment. Although I sometimes can determine how to solve a challenging problem quite quickly (although it usually takes some time, and occasionally an inordinate amount of time!), I don't see anything particularly simple & obvious here. Nonetheless, my initial approach would be something like what is done in the AOPS link in Max Alekseyev's answer. Perhaps later, if this is not fully solved then & I have some time, I'll try to finish it myself. – John Omielan Nov 04 '21 at 06:00
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    @Mike To clarify my last sentence, note Max Alekseyev's answer does fully solve this. I meant determining a simpler solution than using modulo $17043520$, in particular one which can be fully explained & understood within a reasonable size answer. – John Omielan Nov 04 '21 at 07:22
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    @JohnOmielan Thanks for the tip! I'd say it's obvious that there's no solution with $x=2$, but this covers it explicitly and looks cleaner. – Servaes Nov 04 '21 at 07:59
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It is easy to verify computationally that considering the equation modulo $17043520$ implies that it has the only solutions: $$(x,y,z)\in \{ (1, 1, 1),\ (3, 0, 1),\ (5, 2, 2)\}.$$

I refer to the discussion at AOPS on how to find such a modulus. The one I mentioned above may be not the smallest possible, but it does the job.

UPDATE. As established by brute-force, the smallest working modulus here is $128320$.

PS. For the theory behind such equations, see the paper:

J. L. Brenner, L. L. Foster "Exponential Diophantine equations". Pacific J. Math. Volume 101, Number 2 (1982), 263-301. http://projecteuclid.org/euclid.pjm/1102724775

Max Alekseyev
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here is a method I learned from Exponential Diophantine equation $7^y + 2 = 3^x$ and polished. I have answered many questions of the form $A^u - B^v = C$ for given positive integers $A,B,C$

This answer finishes $x=1.$

The following shows that, with $x = 1,$ the largest solution to $9^z - 7^y = 2 $ is $z=y=1.$ First, write $9^z - 9 = 7^y -7$ Next introduce non-negative integers $p,q$ with $$ 9(9^p - 1) = 7 (7^q - 1), $$ ASSUME $p,q \geq 1$ and reach a contradiction. We alternate steps, each is a calculation.

$$ 7 | 9^p -1 \Longrightarrow \; \; 3 | p$$ $$ 9^3 - 1 = 2^3 \cdot 7 \cdot 13 $$ $$ 13 | 7^q -1 \Longrightarrow \; \; 12 | q$$ $$ 7^{12} - 1 = 2^5 \cdot 3^2 \cdot 5^2 \cdot 13 \cdot 19 \cdot 43 \cdot 181 $$

$$ 19 | 9^p -1 \Longrightarrow \; \; 9 | p$$ $$ 9^9 - 1 = 2^3 \cdot 7 \cdot 13 \cdot 19 \cdot 37 \cdot 757 $$ $$ 37 | 7^q -1 \Longrightarrow \; \; 9 | q$$ $$ 7^{9} - 1 = 2 \cdot 3^3 \cdot 19 \cdot 37 \cdot 1063 $$

This number divides $9 (9^p - 1). $ In particular, $27 | 9 (9^p - 1).$ This CONTRADICTS the assumption that $p > 0$

Will Jagy
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This is a PARTIAL answer. What remains to show is that there are not an infinite number of solutions for $x=1$. Assume $x\ge 3$. Then as in Conner's answer, $$2^x = (3^z-7^v)(3^z+7^v).$$

This gives for some positive integer $a$ the equation $$2^a(3^z-7^v)=(3^z+7^v).$$

Rearranging terms gives $$(2^a-1)3^z = (2^a+1)7^v,$$

Note that this implies that $a$ must also be at least 3. So this gives $2^a+1=3^z$. As $a$ is at least $3$ [because one can check that there are no solutions to the above equation for $a \in \{1,2\}$], it follows that $2^a+1=3^z$ and thus in particular $3^z$ must be $1$ mod $8$, and so it follows that $z$ must be even, which gives

$$(3^{z/2}-1)(3^{z/2}+1) = 2^a,$$ which implies both $3^{z/2}-1$, $3^{z/2}+1$, must be powers of $2$. As they differ by only $2$, this is possible only if $z=2$ and $a=3$. So if $x$ is at least $3$, then $v$ must be $1$ and thus $y$ must be $2$. Thus $x$ must be $5$ and $z=2$.

Mike
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    FYI, here's an alternate way to solve for $x \ge 2$ the equation $2^x = (3^z - 7^v)(3^z + 7^v)$. First, note both factors on the right are positive, even and powers of $2$. Thus, $3^z - 7^v = 2^a$ and $3^z + 7^v = 2^b$ with $a \lt b$ and $a + b = x$. Adding the $2$ equations gives $2(3^z) = 2^a + 2^b = 2^a(1 + 2^{b - a})$. Equating the even and odd parts gives $2 = 2^a ; \to ; a = 1$ and $3^z = 1 + 2^{b - a} \implies 3^z - 2^{b - a} = 1$. If $b - a = 1 ; \to ; b = 2$, then $x = 3$ and $z = 1$. ... – John Omielan Nov 04 '21 at 06:13
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    (cont.) Otherwise, with $b - a \gt 1$ and $z \gt 1$, there are proofs here, plus Mihăilescu's theorem can be used, to show the only solution is $z = 2$ and $b - a = 3 ; \to ; b = 4$, so $x = 5$. – John Omielan Nov 04 '21 at 06:14
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Refer to Henri Cohen's Number Theory Volume I: Tools for Diophantine Equations p. $410-411$ and also $461$ for exercise $35$.

For $x = 1$, the equation becomes $2+7^{y}=9^{z} \implies 2+7^{y}=(3^z)^2 \implies (3^{z})^{2}=7^{y}+2.$

On page $410$ we have $6.7$ The Equation:

$y^{2} = x^{n}+t$ for $-100 \leq t \leq -1$ and $n$ even, $t \not\equiv 1 \pmod {8}$, and $t$ squarefree.

By page $419$, the above result is generalized to include odd $n$ as well, showing that each equation satisfying the above has a finite number of solutions, listed in a table in the book.

It is indicated that using the reasoning of the proofs of recent theorems, this result can be extended to $|t| \leq 100.$

It seems a lot of directly related material is in Chapter $6.7$ of this book and the in-text references, as well as the prerequisite material contained within, however it is clear much of it is "left to the reader" to organize as a proof for exercise $35$ which I am personally unable to do myself at this time.

Derek Luna
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    Not sure why I get a downvote, it is a direct reference to a generalization to the equation. One answer just writes it is easy to verify, and the other two don't consider $x=1$, but I get the downvote only. Ok. – Derek Luna Nov 04 '21 at 13:01
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    I upvoted this, because I didn't think it deserved a downvote either. The reference does look interesting. But for questions like these, what is the more satisfying--and expected--for most of us, are reasonably self-contained answers as opposed to referencing another source. – Mike Nov 04 '21 at 13:21
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    there is also a way tyo do $x=1$ with simple calculations, I put an answer. – Will Jagy Nov 05 '21 at 18:43