The only solution in natural numbers of the Diophantine equation $(1) \;\; 125 \cdot 2^n - 3^m = 271$ is $(m,n) = (6,3)$. Proof: Equation (1) implies $5 | 3^m + 1$, yielding $4 | m+2$. Hence $m \geq 2$, which according to equation (1) means $9 | 2^n + 1$, which give us $6 | n - 3$. Thus there exist four non-negative integers $q_1,q_2,r_1,r_2$ s.t. $(2) \;\; m = 12q_1 + 4r_1 + 2, \: r_1 \leq 2,$ and $(3) \;\; n = 12q_2 + 6r_2 + 3, \: r_2 \leq 1$. Combining (1)-(3) we obtain $(4) \;\; (-1)^{r_2+1} - 3^{4r_1+2} \equiv 11 \pmod{13}$. Checking the six combination of $r_1$ and $r_2$, we find that there are two combinations which satisfies congruence (4), namely $(r_1,r_2)=(1,0)$ and $(r_1,r_2) = (2,1)$. Thus we have the following two cases: Case 1: $(r_1,r_2) = (1,0)$. Inserting this in (2)-(3), the results are $(5) \;\; (m,n) = (12q_1+6,12q_2+3)$. Therefore by equation (1) $(6) \;\; x^3 - y^3 = 271$, where $(7) \;\; (x,y) = (5 \cdot 2^{4q_2+1}, 3^{4q_1+2})$. Equation (6) is equivalent to $(x - y)(x^2 + xy + y^2) = 271$, yielding $(x-y,x^2+xy+y^2) = (1,271)$ (since 271 is a prime), which give us $(x,y) = (10,9)$. Consequently $q_1=q_2=0$ by (7), which according to (5) give us $(m,n) = (6,3)$. Case 2: $(r_1,r_2) = (2,1)$. Inserting this in (2)-(3), we obtain $(8) \;\; (m,n) = (12q_1+10,12q_2+9)$. Using the fact that $2^6 \equiv 3^6 \equiv 1 \pmod{7}$ combined with (1) and (8) we obtain $125 \cdot 2^{9} - 3^{10} \equiv 271 \pmod{7}$, i.e. $7 | 4680$. This contradiction implies there is no solution of equation (1). Conclusion: The only solution of equation (1) in natural number is $(m,n) = (6,3)$. q.e.d.

Why do there exist such $q_1,q_2,r_1 and r_2$?

We have $4 | m-2$ and $6 | n-3$. Hence $m-2=4a$ and $n-3 = 6b$. According to the Division Algorithm there exists four non-negative integers $q_1,q_2,r_1,r_2$ s.t. $(a,b) = (3q_1+r_1,2q_2+r_2)$, where $r_1 \leq 2$ and $r_2 \leq 1$, which give us $m = 4a + 2 = 4(3q_1 + r_1) + 2 = 12q_1 + 4r_1 + 2$ and $n = 6b + 3 = 6(2q_2 + r_2) + 3 = 12q_2 + 6r_2 + 3$.

We have $3^m=–1 (mod 16)$, so that $n= 4k+2$. If $n$ ≥ $4$ we will have $9.81^k = 1 (mod 16)$, which contradicts to $81 =1 (mod 16)$. Checking $n$ ≤ $3$ we have the only solution $(6; 3)$.

VicKmath7 wrote: We have 3^m≡ –1 (mod 16), so that n= 4k+2. If n ≥ 4 we will have 9.81^k ≡ 1 (mod 16), which contradicts to 81 ≡ 1 (mod 16). Checking n ≤ 3 we have the only solution (6; 3). sorry but why $3^m\equiv –1 \pmod{16}$

raghoodah1m wrote: VicKmath7 wrote: We have 3^m≡ –1 (mod 16), so that n= 4k+2. If n ≥ 4 we will have 9.81^k ≡ 1 (mod 16), which contradicts to 81 ≡ 1 (mod 16). Checking n ≤ 3 we have the only solution (6; 3). sorry but why $3^m\equiv –1 \pmod{16}$ Move 3*m to the other side

coldheart361 wrote: raghoodah1m wrote: VicKmath7 wrote: We have 3^m≡ –1 (mod 16), so that n= 4k+2. If n ≥ 4 we will have 9.81^k ≡ 1 (mod 16), which contradicts to 81 ≡ 1 (mod 16). Checking n ≤ 3 we have the only solution (6; 3). sorry but why $3^m\equiv –1 \pmod{16}$ Move 3*m to the other side i notice that now thank you

The only solution is (6;3)

Why is it that Equation (1) implies $5 | 3^m + 1$, yielding $4 | m+2$.

$3^m+1$ is periodic because $3^m+1 \equiv 3^{m+4}+1 ( \text{mod 5})$. So we can write $m$ as $4k+r$ when $0 \le r <4$. The only $r$ value is $2$. This implies conclusion

By taking mod 5, RHS=1(mod 5) so 3^m=4(mod 5). Thus, m=4k+2. If n => 4, LHS= -3^m= -3^4k*3^2= -1*9= -9(mod 16) and RHS=15(mod 16), contradiction. So n < 4 or n=1, 2, 3. The only solution is n=3 m=6

Taking $\mod 9$ we get \[ -1 \cdot (2^3)^{n/3} - 3^m \equiv 1 \pmod{9} \]In order for $3^m \equiv 0 \pmod{9}$ we must have $m \ge 2$. Testing out $m=1$ does not yield any solutions so we can safely assume that $3^m \equiv 0 \pmod{9}$. So now we have the congruence \[ -1 \cdot (2^3)^{n/3} \equiv 1 \pmod{9} \]From this we can obviously see that $3\mid n$. Now taking $\mod 10$ we get \[ 5^3 \cdot 2^n-3^m \equiv 1 \pmod{10} \]In order for $5^3 \cdot 2^n \equiv 0 \pmod{10}$ we must have $n \ge 1$. Since $n=0$ is not positive, we have no solutions and again we can safely assume that $5^3 \cdot 2^n \equiv 0 \pmod{10}$. So we have the congruence \[ -3^m \equiv 1 \pmod{10} \]So we have $m=4b+2$. Last mod I promise. Finally we take $\mod 16$ and we get the congruence \[ 5^3 \cdot 2^{3a}-3^{4b+2} \equiv -1 \pmod{16} \]In order for $5^3 \cdot 2^{3a} \equiv 0 \pmod{16}$ we must have $a>=2$. Testing out $a=1 \implies n=3$ satisfies the equation with the ordered pair $(m,n)=(6, 3)$. So we can safely assume that $5^3 \cdot 2^{3a} \equiv 0 \pmod{16}$. So we have the congruence, \[ -3^{4b+2} \equiv -3^{4b} \cdot 9 \equiv -(81) \cdot 9 \equiv -1 \cdot 9 \equiv -9 \not \equiv -1 \pmod{16} \]So we have no other solutions. Thus our only solution is the ordered pair $(m,n)=(6,3)$.

taking mod 5 we get that 3^m = 4 (mod 5) => m = 4k +2 now suppose that n>=4 => 2^n * 125 = 0 (mod 16) => 3^m = 1 (mod 16) => m=4k and this is contradiction. so n<=3 and only suitable solution is n=3, m =6.