Peter wrote: Solve the equation $ 2^x - 5 = 11^{y}$ in positive integers. $ x \geq 2$ We have: $ 11^y +5 \equiv (-1)^y+1 \equiv 2^x \equiv 0(mod 4)$ So $ y=2k+1$ with $ k \in N$ $ 2^x \equiv 5+11^y \equiv (-1)+(-1)^y \equiv -2 \equiv 1(mod 3)$ So $ x=2z$ with $ z \in N$ Then $ 4^z-5=11^y$ $ 4^z \equiv 11^y+5 \equiv 1(mod 5)$ =>$ z=2t$ =>$ 16^t=11^y+5$ So $ 11^y \equiv -5(mod 16)$ Otherhand $ 11^y \equiv (-5)^y (mod 16)$ So $ y=1$ then $ x=4$ I hope my solution is true !

I follow completely until the second last line: how does it follow $ y=1$?

i don't think the solution isn't true because you made a mistake : $ ( - 5)^y \equiv - 5 (mod16)\Rightarrow y = 1$ it's not true !!!

Peter wrote: Solve the equation $ 2^x - 5 = 11^{y}$ in positive integers. It's easy t prove that $ x \vdots 4$ and y is odd.So $ x=4x';y=2y'+1$. And $ 2^{2x'}=a;11^{y'}=b$.We have: $ a^2-11b\&2=5. (1)$ It's is Pell Equation.All solutions of (1) are given by(note that we have$ a \vdots 2$ ) \[ \left\{ \begin{array}{l} a_1 = 4,b_1 = 1 \\ a_{n + 1} = 10a_n + 33b_n \\ b_{n + 1} = 3a_n + 10b_n \\ \end{array} \right.\] But $ a_j (j \ge 1)$is odd.So we have a solution $ a=4,b=1.$ Thus,we infer that $ x=4,y=1$

tronghieu wrote: Peter wrote: Solve the equation $ 2^x - 5 = 11^{y}$ in positive integers. It's easy t prove that $ x \vdots 4$ and y is odd.So $ x = 4x';y = 2y' + 1$. And $ 2^{2x'} = a;11^{y'} = b$.We have: $ a^2 - 11b \& 2 = 5. (1)$ It's is Pell Equation.All solutions of (1) are given by(note that we have$ a \vdots 2$ ) \[ \left\{\begin{array}{l} a_1 = 4,b_1 = 1 \\ a_{n + 1} = 10a_n + 33b_n \\ b_{n + 1} = 3a_n + 10b_n \\ \end{array} \right. \] But $ a_j (j \ge 1)$is odd.So we have a solution $ a = 4,b = 1.$ Thus,we infer that $ x = 4,y = 1$ Sorry I don't think you're right.I think we can't conclude that $ a_j (j \ge 1)$ right?

hxy09 wrote: Sorry I don't think you're right.I think we can't conclude that $ a_j (j \ge 1)$ right? Thanks hxy09. I'm not careful in some calculation . This is my solution. Consider modulo $ 5$, $ 2^x \equiv 1(\bmod 4) \Rightarrow x \vdots 4.$ and modulo $ 16$, we have $ y \equiv 1(\bmod 4).$ If $ y = 1$ then $ x = 4$. We'll prove $ (1;4)$ is unique solution. In fact, if $ y \ge 2$ , with some check calculation, we can prove $ x \ge 6$. Thus, $ 2^x \vdots 2^6$ . Morever, we have: $ 2^x = 11^y + 5 = (2^3 + 3)^y + 5 = (2^3 )^y + \left( {\begin{array}{*{20}c} y \\ 1 \\ \end{array}} \right)(2^3 )^{y - 1} .3 + ... + \left( {\begin{array}{*{20}c} y \\ {y - 1} \\ \end{array}} \right)2^3 .3^{y - 1} + + 3^y + 5.$ So $ 8y.3^{y - 1} + 3^y + 5 \vdots 2^6 .$ Consider $ y \equiv 1(\bmod 2^3 )$ then $ 8y \equiv 8(\bmod 2^6 )$. So $ 8y.3^{y - 1} + 3^y + 5\equiv 3^{y + 1} + 5(\bmod 2^6 )$. However, $ \left( {\frac {{ - 5}}{{2^6 }}} \right) = - 1$. That gives contracdition. Thus, $ y \equiv 5(\bmod 2^3 ). (1)$ Setting $ x = 4x' ( x' \in N, x' > 1)$. Rewrite our equation: $ 16(16^{x' - 1} - 1) = 11(11^{y - 1} - 1).$ Because $ gcd(16,11) = 1,$ so $ 16^{x' - 1} - 1\vdots1$1. That gives $ x' - 1\vdots5.$So $ 16^{x' - 1} - 1\vdots 16^5 - 1\vdots 41$. From that, we have $ 11^{y - 1} - 1 \vdots 41.$ So$ y - 1 \vdots 40. (2)$ From $ (1)$ and $ (2)$ we have the contracdition. That finishes proff.

Very nice your solution I am impressed by your setting $ 11=2^3+3$ This is a very efficient way!

thanks hxy09. I've edited some mistakes in typing.

tronghieu wrote: Consider $ y \equiv 1(\bmod 2^3 )$ then $ 8y \equiv 8(\bmod 2^6 )$. So $ 8y.3^{y - 1} + 3^y + 5\equiv 3^{y + 1} + 5(\bmod 2^6 )$. However, $ \left( {\frac {{ - 5}}{{2^6 }}} \right) = - 1$. That gives contracdition. Thus, $ y \equiv 5(\bmod 2^3 ). (1)$ another way to prove this is by lifting the exponent in fact,we have $v_2(y-1)=2$ tronghieu wrote: $ 11^{y - 1} - 1 \vdots 41.$ So$ y - 1 \vdots 40. (2)$ to conclude this we must prove that $ord_{41}(11)=40$ . Am I missing something?