Find all integer pair $(m,n)$ such that $7^m=5^n+24$.
Problem
Source: KJMO 2018 p7
Tags: number theory, Diophantine equation
Solar Plexsus
27.07.2019 09:12
We have given the Diophantine equation $(1) \;\; 7^m = 5^n + 24$. According to equation (1) $(-1)^m \equiv 7^m = 5^n + 24 \equiv 1^n = 1 \pmod{4}$, implying $m$ is even. Futhermore $1 \equiv 7^m = 5^n + 24 \equiv (-1)^n \pmod{3}$, yielding $n$ is even. Hence $(m,n) = (2s,2t)$ ($s,t \in \mathbb{N}$), which inserted in equation (1) result in $(2) \;\; (7^s - 5^t)(7^s + 5^t) = 24$. The fact that $2 \mid 7^s \pm 5^t$ combined with equation (2) give us $(7^s - 5^t,7^s + 5^t) = (2,12)$ or $(7^s - 5^t,7^s + 5^t) = (4,6)$. Consequently $(7^s,5^t) = (7,5)$ or $(7^s,5^t) = (5,1)$, which means $s=t=1$. Conclusion The only solution of equation (1) is $m=n=2$.
AlastorMoody
08.08.2019 21:34
KJMO 2018 P7 wrote: Find all integer pair $(m,n)$ such that $7^m=5^n+24$. Solution: Note that: $\text{ord}_4 (7)=2$ and $\text{ord}_6 (5)=2 \implies m=2m_1$ and $n=2n_1$ $$7^m=5^n+ ~ 24 \implies (7^{m_1}+ ~5^{n_1})(7^{m_1}- ~5^{n_1})=24 \implies (m_1,n_1) \equiv (1,1) \implies \boxed{(m,n)\equiv (2,2)}$$
Mahdi_Mashayekhi
10.02.2022 19:37
$7^m = (-1)^m = 5^n + 24 = 1^n (mod 4)$ so $m$ is even so $ m = 2x$. $7^m = 1^m = 5^n + 24 = 2^n (mod 3)$ so $n$ is even so $ n = 2y$. $7^m - 5^n = 24$ so $((7^x - 5^y)(7^x + 5^y)) = 24$. $(7^x - 5^y,7^x + 5^y) = (1,24) , (2,12) , (3,8) , (4,6) $ and it's obvious $(7^x - 5^y,7^x + 5^y)$ = $(2,12)$ is the answer so $x = y = 1$ so $m = n =2$.
Math4Life2020
11.02.2022 01:29
KJMO 2018/7 wrote: Find all integer pair $(m,n)$ such that $7^m=5^n+24$.
Firstly, take (mod 6) to get 1 = 5^n (mod 6) -> 2|n.
Now, take (mod 8) to get 7^m = 1 (mod 8) -> 2|m.
Take (m,n)=(2a,2b).
Then, we have: 7^(2a)-5^(2b)=24=(7^a+5^b)(7^a-5^b).
By parity, we have that 2 divides both of these factors, and that the first is greater than the second.
Case 1. 7^a+5^b=12, 7^a-5^b=2 -> (a,b)=(1,1) -> (m,n)=(2,2).
Case 2. 7^a+5^b=6, 7^a-5^b=4 -> 7^a=5 -> impossible.
Thus (m,n)=(2,2) is the only solution in integers.
strong_boy
06.11.2022 18:14
$7^{m} = 5^{n}+24$ . By $mod 4$ we can see LHS in $mode 4$ is $(-1)^m$ and RHS is $1$ . Then we get $m=2t$ . By $mode 3$ we can see LHS in $mode 3$ is $1$ . Then we get $n=2u$ . Then we can write equation as this : $$24 = (7^{t}+5^{u})(7^{t}-5^{u})$$ Now it is easy to check .....
JH_K2IMO
17.12.2024 11:15
7^m=5^n+24. 7^m≡1 , 5^n≡(-1)^n, 24≡0. 1≡(-1)^n (mod 3). Thus, n=2a. 7^m≡(-1)^m, 5^n≡1,24≡0. (-1)^m≡1 (mod 4). Thus, m=2b. Let's substitute into the equation. 49^b=25^a+24. (7^b+5^a)(7^b-5^a)=24. {(7^b+5^a) ,(7^b-5^a)}=(24,1),(12,2),(8,3),(6,4). The only possible ordered pair among the above is (12, 2). a=1,b=1. m=2,n=2. ∴The answer is (2,2).
wizixez
17.12.2024 13:09
See Azerbaijan JNMO 2020 P5 for a harder version where 24 is n!