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$k\in N^+$.

Does there exists positive integer solutions (x,y) of Pell equation $x^2-2y^2=8k+1$for any $k$?

If $8k+1$ is a prime, there exists positive integer solutions.

See Here.

See Here.

But for other numbers,I have tried several ways to solve the equation and finally failed.

Any ideas?

A waiting the reply.

qsa
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2 Answers2

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No, there does not exist a solution for any $k$. Here is a counterexample. The equation has no solution for $k=4$, i.e., $$x^2-2y^2=33$$ has no integer solution. The same holds for $k=7$. If $8k+1$ is a perfect square, there is always a solution with $y=0$.

Dietrich Burde
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If an odd number $n$ is divisible only by primes $p \equiv \pm 1 \pmod 8,$ then it can be expressed in integers as $n = x^2 - 2 y^2.$

For any prime $q \equiv \pm 3 \pmod 8,$ there is no expression for $q$ as $u^2 - 2 v^2.$ With an odd exponent $w = 2k+1,$ there is also no expression for $q^w$ as $u^2 - 2 v^2.$ An even exponent here produces a square so that can be expressed with $v=0$

Put together, an odd number $n$ can be expressed if and only if the exponent of prime $q |n$ is even, whenever $q \equiv \pm 3 \pmod 8.$

In Dietrich's example, note how $33 = 3 \cdot 11,$ while $3 \equiv 3 \pmod 8$ and $11 \equiv 3 \pmod 8$

Will Jagy
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  • How to prove : For any prime $q \equiv \pm 3 \pmod 8,$ there is no expression for $q$ as $u^2 - 2 v^2. – qsa Aug 04 '19 at 00:32
  • @qsa the values of $t^2 \equiv 0,1,4 \pmod 8.$ Thus the values of $-2 s^2 \equiv 0, 6 \pmod 8.$ Put them together, $u^2 - 2 v^2 \equiv 0,1,2,4,6,7 \pmod 8.$ I can recommend the recent book https://bookstore.ams.org/mbk-105/ – Will Jagy Aug 04 '19 at 00:39
  • $33=3*11$.Let $q_1=3$ and $q_2=11$.

    $q_1,q_2 \equiv 3 \pmod 8$

    How to prove:there is no expression for $q_1q_2$ as $u^2 - 2 v^2.$

     – qsa Aug 04 '19 at 00:49