Peter wrote: Solve the equation $ x^2 + 7 = 2^n$ in integers. Here is my solution We have: $ x^2=2^n-7 \equiv (-1)^n-1(mod 3)$ If $ n=2k+1;k \in Z$ then $ x^2 \equiv -2(mod 3)$,wrong So $ n=2k$ with $ k \in Z$ Then $ 2^{2k}-x^2=7$ <=>$ (2^k-x^2)(2^k+x^2)=7$ =>$ 2^k+x^2=7 and 2^k-x^2=1$ So $ 2.2^k=k=>k=2$ =>$ n=4$=>$ x^2=9$=>$ x =\pm 3$

What's wrong with $ x^2\equiv -2\pmod3$? Schmu

Nothing is wrong with it. There are actually such solutions: $ x=1, x=5, ...$. So please try again.

Peter wrote: Solve the equation $ x^2 + 7 = 2^n$ in integers. This one is famous,but it seems there are no elementary proof.

Posted. This equation goes back to Ramanujan and the standard solution due to Nagell involves a fair amount of work in $ \mathbb{Z} \left[ \frac{1 + \sqrt{-7}}{2} \right]$.

Hmm... not really a PEN-problem if that's what it needs to solve this.

Peter wrote: Hmm... not really a PEN-problem if that's what it needs to solve this. As one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it ... Also this problem was used at that Olympiad,although nobody was able to find a proof without using Gaussian numbers during olympiad and as I can see at ML elementary solution wasn't discovered for a long time too. This remark gives us hope that there must exist "almost elementary" proof and let's just wait until somebody... Unfortunately Teacher who gave us this problem can not find the place where he saw elementary solution and I have a sense that he maybe didn't even try to search Best regards.

Erken wrote: Peter wrote: Hmm... not really a PEN-problem if that's what it needs to solve this. As one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it ... Maybe the person putting this in the test didn't know it was so difficult, or had a wrong solution in mind (like the few posted on ML)? The fact that this is a famous problem that Ramanujan couldn't solve, strongly suggests that this is NOT a pre-olympiad level problem.

Peter wrote: Solve the equation $x^2 +7=2^n$ in integers. $X^2 +7=2^n$ $X^2+16-9 =2^n$ $(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$ We see $x =2k+1$ $=>(k+2)(k-1)=4(2q-1)$ Observe $(k+2,k-1) =1$ for being odd or even, And also $(4,2q-1) =1$ $k+2=4$ or $k-1=4$ $=> k=2$ or $k=5$ $X=5$ or $x=11$ For $n=4, x=3$ For $n<4$, we may check the values of n. $(x,n) =(1,3),(3,4),(5,5),(11,7)$

$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$ We see $x =2k+1$ $=>(k+2)(k-1)=4(2q-1)$ It should be $8(2^q-1)$ . How did you come up with $4(2q-1)$?

reveryu wrote: $(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$ We see $x =2k+1$ $=>(k+2)(k-1)=4(2q-1)$ It should be $8(2^q-1)$ . How did you come up with $4(2q-1)$? Consider (X+3)(X-3)=2(k+2)2(k-1)=4(k+2)(k-1) But you're right about 2q-1.it should be 2^q-1 That was a perfect solution

venkats wrote: $=>(k+2)(k-1)=4(2q-1)$ Observe $(k+2,k-1) =1$ for being odd or even, And also $(4,2q-1) =1$ $k+2=4$ or $k-1=4$ Counter-example: $2 \cdot 15= 6\cdot 5$?

venkats wrote: Peter wrote: Solve the equation $x^2 +7=2^n$ in integers. $X^2 +7=2^n$ $X^2+16-9 =2^n$ $(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$ We see $x =2k+1$ $=>(k+2)(k-1)=4(2q-1)$ Observe $(k+2,k-1) =1$ for being odd or even, And also $(4,2q-1) =1$ $k+2=4$ or $k-1=4$ $=> k=2$ or $k=5$ $X=5$ or $x=11$ For $n=4, x=3$ For $n<4$, we may check the values of n. $(x,n) =(1,3),(3,4),(5,5),(11,7)$ $(k+2,k-1) =3$ for k=1 so k-1=4*3 or 4 at most but anyways nice solution.