$k\in N^+$.
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.
But for other numbers,I have tried several ways to solve the equation and finally failed.
Any ideas?
A waiting the reply.
On
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 $11 \equiv 3 \pmod 8$
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 $y=0$.