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I'm working on this problem I came across on the internet but I have no solution yet. The problem states: Find all prime numbers p that are such that $p+1=2x^2$ and $p^2+1=2y^2$ where x and y are integers.

Edit: to which point I have reached.
I wrote $p=2p_1+1$. After this the second equation becomes $2p_1^2+2p_1+1=y^2$ which is rewritten as $p_1^2+x^4=y^2$ in which we only have to find Pythagorean triples.

3 Answers3

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Suppose wlog $x,y >0$

Substract initially the two equations to get $p(p-1)=2(y-x)(y+x)$
If $p \mid (y-x) \Rightarrow p \leq y-x \Rightarrow p-1 \geq 2(y+x)$ so we can see that $y-x > 2(y+x) \Rightarrow y+3x<0$ contradiction.

Therefore $p \mid (x+y) \Rightarrow p\leq x+y \Rightarrow p-1\geq 2(y-x)$ so $x+y \geq 2(y-x) \Rightarrow y<3x \Rightarrow y^2<9x^2 \Rightarrow p^2+1<9(p+1)$

We now just have to solve $p^2-9p-8<0 $ from which we deduce that $p<10$ and by checking possible values we get $p=7$.

miracle173
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    This is a really elegant use of divisibility (i.e., the FTA) plus inequalities. Nicely done! – Kieren MacMillan Aug 23 '14 at 14:24
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    I agree! Approach is clever and straightforward. 1+ – Markus Scheuer Aug 23 '14 at 14:33
  • It would be very interesting to see if this proof could be adapted to prove the more general [and true] case where $p=n$ is any positive integer. – Kieren MacMillan Aug 23 '14 at 15:58
  • What do you mean by true? What's true for p=n? –  Aug 23 '14 at 19:57
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    Let $n$ be any positive integer, prime or otherwise, such that the system of equations $n+1=2x^2$ and $n^2+1=2y^2$ has an integer solution $(x,y)$. Then $n=1$ or $n=7$. Your proof in the case of $p$ prime might be applicable to this theorem, but the method would have to be modified somewhat to account for the prime factorization of $n$. – Kieren MacMillan Aug 24 '14 at 16:43
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The general solution to diophantine equations using pythogorean triples is:

$$x^2 + y^2 = z^2$$

And $$(x, y, z) = (2rs, r^2 -s^2, r^2 +s^2)$$

You can use this to find $p_1$ and $p$

EDIT

Notice actually that in the second equation we can rewrite to get:

$$\large p^2 - 2y^2 = -1$$

This is an example of a Pell Equation, where $D = 2$

You can use the continued fraction expression for the $\sqrt{2}$ to find a fundamental solution:

$$\large \sqrt{2} = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2 +...}}}$$

And so the convergents are:

$$\frac{1}{1}, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, ...$$

From the convergents, we can see that the first solution is when $p = 7$, and $y = 5$ (the first solution is technically, $(1,1)$, but $1$ isn't prime).

By evaluating the convergents, we can see that the 1st, 3rd, 5th, etc. are solutions to the Pell Equation.

You can more about it here

hope this helped.

Varun Iyer
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  • I know that but still I can't find the actual values of p –  Aug 10 '14 at 16:45
  • Use small value of r and s to find possible prime values – Varun Iyer Aug 10 '14 at 16:47
  • @Matheo I found a better way to solve it. hope this helps you – Varun Iyer Aug 10 '14 at 17:03
  • Two observations. First, the sequence of solutions to the Pell equation, $1,7,41,\ldots$ satisfies the recurrence $p_{n+2}=6p_{n+1}-p_n$. Second, the problem is now to find which elements of this sequence are simultaneously prime, and of the form $2x^2-1$. This does not seem easy to do, but perhaps there is some trick here. In any case, it is likely that $p=7$ is the only solution. – Andrew Dudzik Aug 10 '14 at 17:21
  • @you-sir-33433 yes I realize that now, but its a huge step forward for finding the solutions for him, though its still a tenous pain to find which are prime and satisfy both equations too... – Varun Iyer Aug 10 '14 at 17:22
  • I'm very curious about a solution, however much pain is involved. This is similar to the problem of classifying square Fibonacci numbers, which can be done with elementary methods (though the same tricks do not work here). – Andrew Dudzik Aug 10 '14 at 17:26
  • @you-sir-33433 I am very curious too. This is as far as I could get. I only know elementary solutions to these equations. I'm not very proficient in number theory :(. – Varun Iyer Aug 10 '14 at 17:27
  • Has it been shown (or is it trivial to show) that (3,4,5) are the only Pythagorean numbers such that the smallest two differ by 1? –  Aug 10 '14 at 18:29
  • As it turns out, it isn't true :/ –  Aug 10 '14 at 18:32
  • No there is 9 40 41 – Varun Iyer Aug 10 '14 at 18:38
  • @VarunIyer In your example the smallest two don't differ by $1$. – coffeemath Aug 12 '14 at 22:52
  • True but still there are other examples. –  Aug 12 '14 at 23:11
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$$p+1=2x^2,p^2+1=2y^2\\ 2y^2-1=(2x^2-1)^2\\ y^2=2x^4-2x^2+1\\ y^2=2x^2 (x-1)(x+1)+1$$ Suppose $y=2m+1,$ then $$y^2=4m(m+1)+1.$$ Hence $$2m(m+1)=x^2 (x-1)(x+1).$$ $$\dfrac{m(m+1)}{2}=\dfrac{x(x-1)}{2}\times\dfrac{x(x+1)}{2}$$ I have no idea to continue this. But I think find the pairs triangle numbers who can satisfy the equation $$\color{Green}{T_m=T_{x-1}T_x}$$ would be a help to solve tyhis problem.

Bumblebee
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