Peter wrote: Solve the equation $ 7^x - 3^y = 4$ in positive integers. If $ x=1=>y=1$ or if $ y=1$ =>$ x=1$ If $ x \geq 2;y \geq 2$ *Case 1:$ x=2k$ with $ k \in Z+$ We have $ 3^{2k}+4=7^x$ =>$ 9^k+4=7^x$=>$ 7^x \equiv 1(mod 4)$ =>$ x =2m$ with $ m \in Z^+$ Then we have $ (7^m-3^k)(7^m+3^k)=4$=>$ 7^m-3^k=1 and 7^m+3^k=4$ =>$ 2.7^m=5$=>$ m$ isn't integer number *Case 2:$ x=2k+1$ Then $ 3^{k+1}+4=7^x$=>$ 3.9^k+4=7^x$ =>$ 7^x \equiv 3(mod 4)$ wrong because $ 7^x \equiv (-1)^x (mod 4)$ So $ (x;y)=(1;1)$

chien than wrote: Peter wrote: Solve the equation $ 7^x - 3^y = 4$ in positive integers. If $ x = 1 = > y = 1$ or if $ y = 1$ =>$ x = 1$ If $ x \geq 2;y \geq 2$ *Case 1:$ x = 2k$ with $ k \in Z +$ We have $ 3^{2k} + 4 = 7^x$ =>$ 9^k + 4 = 7^x$=>$ 7^x \equiv 1(mod 4)$ =>$ x = 2m$ with $ m \in Z^ +$ Then we have $ (7^m - 3^k)(7^m + 3^k) = 4$=>$ 7^m - 3^k = 1 and 7^m + 3^k = 4$ =>$ 2.7^m = 5$=>$ m$ isn't integer number *Case 2:$ x = 2k + 1$ Then $ 3^{k + 1} + 4 = 7^x$=>$ 3.9^k + 4 = 7^x$ =>$ 7^x \equiv 3(mod 4)$ wrong because $ 7^x \equiv ( - 1)^x (mod 4)$ So $ (x;y) = (1;1)$ Wrong!

Maybe you can give the correct solution?

ali666 wrote: let $ x=a+1 , y=b+1$ we have:$ 3(3^{a} - 1) = 7(7^{b} - 1)$ therefore $ 7|3^{a} - 1$ it give $ 6|a$(because $ ord_{7}^{3} = 6$) $ 13|3^{6} - 1,3^{6} - 1|3^{a} - 1\Rightarrow13|3^{a} - 1\Rightarrow13|7^{b} - 1$ $ ord_{13}^{7} = 12$therefore $ 12|b$ $ 7^{12} - 1 = (7^{6} + 1)(7^{3} + 1)(7^{2} + 7 + 1)(7 - 1) = = (7^{6} + 1)(7^{3} + 1)(19)(9) \Rightarrow9|7^{12} - 1,7^{12} - 1|7^{b} - 1\Rightarrow9|7^{b} - 1\Rightarrow 9|3(3^{a} - 1)\Rightarrow{a = 0}\Rightarrow{b = 0}\Rightarrow{x = 1,y = 1}$

See also https://artofproblemsolving.com/community/c6h48431

There is an easier way to prove that! Lemma : If ab =cd, and (a, b) = (c, d) = (a, d) =1 , then : a = c and b = d. (a,b,c,d ≠ 0) That's really easy! [ a|cd , (a,d)=1 => a|c. (1) c|ab , (a,b)=1 => c|a or c|b if c|a then the lemma is proved. if c|b , from (1) we know a|c so a|b, but (a,b) =1 and it's wrong. ] Now let's get back to our problem! Let x=b+1 and y=a+1. We know : 3(3^b - 1) = 7(7^a - 1) (1) Besides, we know : ( 3 , 3^b - 1 ) = ( 7 , 7^a - 1) = (3,7) = 1. So: 3 = 7^a -1 and that's impossible ! So the equation (1) is 0=0 , which means a = b = 0. So x = y = 1!

mhdr_haamiii wrote: There is an easier way to prove that! Lemma : If ab =cd, and (a, b) = (c, d) = (a, d) =1 , then : a = c and b = d. (a,b,c,d ≠ 0) That's really easy! [ a|cd , (a,d)=1 => a|c. (1) c|ab , (a,b)=1 => c|a or c|b if c|a then the lemma is proved. if c|b , from (1) we know a|c so a|b, but (a,b) =1 and it's wrong. ] Now let's get back to our problem! Let x=b+1 and y=a+1. We know : 3(3^b - 1) = 7(7^a - 1) (1) Besides, we know : ( 3 , 3^b - 1 ) = ( 7 , 7^a - 1) = (3,7) = 1. So: 3 = 7^a -1 and that's impossible ! So the equation (1) is 0=0 , which means a = b = 0. So x = y = 1! $c|ab , (a,b)=1 => c|a or c|b$ This is wrong. consider $c=6 ,a=2 ,b=3$

Clearly $x = y = 1$ provides a solution. We claim that there are no others. When $y \ge 2$, we get $3^y \pmod{234} \in \{9, 27, 81\} = A$, while $7^x \pmod{234} \in \{1, 7, 49, 109, 61, 193, 181, 97, 211, 73, 43, 67\} = B$ for all $x \in \mathbb{N}$. Since $(A + 4) \cap B = \emptyset$, there is no solution for $y \ge 2$.