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My problem is related to the equation from above. It actually is a very particular one. I noticed that for every positive integer $n$ there's ONE SINGLE solution $(x_1,y_1)$ so that $x_1$ and $y_1$ are ODD positive integers ( I didn't prove it, I tested it with a program through roughly 30 tests). If that's true , then how can I prove it? I tried to make a proof by absurd reduction ( assuming the contrary of sentence , that is , there could be at least one more solution for some values of n but not necessarly any n and run into a contradiction) which didn't work, and personally I can't "feel" why it can't be more than one solution. I hope you could help me to prove it. Thanks in advance! P.s.: $n>=3$ .

Anonymus
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  • How do you solve $2=7x^2+y^2$ ? Here, $n=1$,. certainly a positive integer. For larger exponents, you can build a solution by starting with a solution for $n=2,3$ and multiplying by $2^{2k}$ for appropriate $k$. – lulu Jul 07 '17 at 16:53
  • Ah, I missed the condition that $x,y$ be odd. My trick doesn't work with that rule. – lulu Jul 07 '17 at 16:55
  • @lulu OP is restricting allowed solutions to odd positive integers, so the solutions obtained by multiplying by $2^{2k}$ are irrelevant. Also, this pattern has to start at $n=3$ since there is no solution to $4=7x^2+y^2$ either. – Mark Fischler Jul 07 '17 at 16:57
  • @MarkFischler Yeah, I missed the "odd" condition (pointed that out in my second comment). – lulu Jul 07 '17 at 16:59
  • Possible duplicate of Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers – lulu Jul 07 '17 at 17:19
  • @lulu the equation is the same but my question is NOT about the way I can prove that for any integer n>=3 there is a solution of the equation, (x,y), with x and y odd and positive integers, it's about the way I can prove that there's EXACTLY one solution for any integer n>=3. (don't know how to use bold) – Anonymus Jul 07 '17 at 17:27
  • Understood. I'll retract the close vote accordingly, but I'll leave up the duplicate citation, as I expect that the solutions provided there will provide guidance toward a proof of uniqueness. – lulu Jul 07 '17 at 17:30
  • Indeed, the first posted solution to the earlier question includes a proof of uniqueness (Note: I did not check that argument carefully, but it looked sensible on a quick read). – lulu Jul 07 '17 at 17:32

1 Answers1

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2^n ≡ y^2 mod 7 ⇒ y^2 < 7

It can be seen that y^2 can only have values 1, 2 and 4. Only 1 and 4 are acceptable if y is integer, that is y= (+ or -) 1 or y =(+ or -) 2. With these values of y some powers of n give integers for x, for example y=2, x =6 for n=8 or x=y=2 for n=5. This is the reason for existing only one single x and y for every n.

sirous
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