$2^t=3^x \cdot 5^y+7^z$

Question

Solve $2^t=3^x \cdot 5^y+7^z$ in positive integers.

My first thought was that I could solve them problem by using some kind of modular arithmetic. I tried modulus up to $20$. And the only thing I found is that $t \equiv 0 \pmod 2$. I found that when I checked modulus $3$.

$2^t \equiv 1 \text{ or } 2 \pmod 3$;

$3^x \cdot 5^y \equiv 0 \pmod 3$;

$7^z \equiv 1 \pmod 3$;

$\implies 2^t \equiv 1 \pmod 3 \implies t \equiv 0 \pmod 2$

Something else that I thought of was to use Euler's Theorem, but I couldn't make it work.

I will appreciate even the smallest help as I feel like I have hit a brick wall.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Oct 16 '22 at 10:35
Please, learn to use MathJax to type the math in your questions. — jjagmath, Oct 16 '22 at 10:42
@KevinDietrich Use $a \equiv b \pmod m$ to get $a\equiv b \pmod m$ with the correct spacing. — jjagmath, Oct 16 '22 at 10:52
Please, avoid the use of unusual characters, they won't render properly in some devices. Use MathJax — jjagmath, Oct 16 '22 at 11:05
Unfortunately, $\Bbb N$ doesn't have a standard definition, since some people include $0$ as a natural number and some people don't. Which convention are you considering? — jjagmath, Oct 16 '22 at 11:29
Once you know that $t$ is even, consider the equation modulo $5$ — jjagmath, Oct 16 '22 at 11:57
"My first thought was that I could solve them problem by using some king of modular arithmetic." I suppose Gauss or Fermat would have been good. — Oscar Lanzi, Oct 16 '22 at 16:09
In positive integers , the only solution I found so far is $2^6=3\cdot 5+7^2$ — Peter, Oct 16 '22 at 16:52
@Peter I checked with a computer for all positive numbers up to 30 and the solution that you have noted is the only one I found as well. — 9_SOS_9, Oct 16 '22 at 17:07
Does this answer your question? Find all integer numbers $k$,$l$,$m$,$n$ for which $3^k$ • $5^l$ + $7^m$ = $2^n$ - found using an Approach0 search. Another basic duplicate found is Find $a, b, c, d \in \mathbb{Z}$ such that $2^a=3^b5^c+7^d$. — John Omielan, Oct 16 '22 at 18:06
@Squaller_ Welcome to Math SE. FYI, the search also found multiple AoPS threads, e.g., Number theory, Number theory nice problem, JBMO 2012 Problem 4, Famous problem, Number Theory, Problem., etc. — John Omielan, Oct 16 '22 at 18:11
Sad to see it's a dupe from so many places. I guess I wasted my efforts then figuring it out and answering it here — Mike Daas, Oct 16 '22 at 18:29

score 1 · Accepted Answer · answered Oct 16 '22 at 17:49

As already noted, reducing the equation modulo $3$ yields that $2^t \equiv 1 \mod 3$ and as such, $t$ must be even. If we then reduce the equation modulo $5$, it follows that $2^t \equiv 7^z \equiv 2^z \mod 5$. This means that $t - z$ is divisible by $4$. In particular, $z$ is even too. Now write $t = 2a$ and $z = 2b$. We may then rewrite the equation as $$ (2^a - 7^b)(2^a + 7^b) = 3^x 5^y. $$ Unique factorisation and the second factor being positive then implies the existence of $x_1,x_2 \geq 0$ with $x_1 + x_2 = x$ and $y_1,y_2 \geq 0$ with $y_1 + y_2 = y$ satisfying $$ 2^a - 7^b = 3^{x_1}5^{y_1} \quad \text{and} \quad 2^a + 7^b = 3^{x_2}5^{y_2}. $$ Adding these equations yields that $$ 2^{a+1} = 3^{x_1}5^{y_1} + 3^{x_2}5^{y_2}. $$ If both $x_1, x_2 \geq 1$, divisibility by $3$ yields a contradiction. Hence either $x_1 = 0$ or $x_2 = 0$. Similarly, we must have either $y_1 = 0$ and $y_2 = 0$. Note that we cannot have $x_2 = y_2 = 0$ because $2^a + 7^b > 1$. There are thus three cases left to consider.

Suppose that $x_1 = y_1 = 0$, so in particular $2^a - 7^b = 1$. Suppose that $a \geq 4$ and reduce the equation modulo $16$. Then it follows that $7^b \equiv -1 \mod 16$, but one can quickly check that this yields no solutions as $7^2 \equiv 1 \mod 16$. We reduce to $a \leq 3$ and quickly find that only $a = 3$ and $b = 1$ yields a valid solution. This results in the solution $$ 2^6 = 3 \cdot 5 + 7^2 $$ found in the comments.

Now suppose that either $x_1 = y_2 = 0$ or $x_2 = y_1 = 0$. Then depending on the case, it either holds that $$2^a - 7^b = 5^y \quad \text{or} \quad 2^a - 7^b = 3^x. $$ To avoid the left hand side being negative, we may assume that $a \geq 3$. Now reduce the equation modulo $8$ to find that $\pm 1 \equiv 5^y \mod 8$ or $\pm 1 \equiv 3^x \mod 8$. This readily implies that $y$ resp. $x$ must be even.

However, we must in both cases also have that $2^{a+1} = 3^x + 5^y$. Reducing this modulo $8$ then yields that $0 \equiv 3^x + 1 \mod 8$ or $0 \equiv 1 + 5^y$ but this is a contradiction in both cases.

$2^t=3^x \cdot 5^y+7^z$

1 Answers1