All the variables I deal with are integers. Given $$q \geq 2, |k| \geq 2$$, impose a constraints that we have an integer $$p$$ such that $$q^2=1+pk=1\pmod k.$$ For example, $$3^2=1+1\cdot8, 5^2=1+3\cdot8.$$ Now

I suspect that given the triple $$(q,k,p)$$, then at least one of the following is true.

(1) there is a shift integer $$u$$ such that $$(q+ku)^2=1+p_1k$$ where $$p_1$$ is a square

(2) there is a shift integer $$u$$ such that $$(-q+ku)^2=1+p_1k$$ where $$p_1$$ is a square.

For example, given $$q=5, k=8, p=3$$. I can find $$u=1$$ so that the second case $$-q+ku=-5+8(1)=3$$ and then $$3^2=1+8\cdot1$$ now $$1$$ is a square. Notice that if original $$p$$ is a square then we are done or $$k=\pm 1$$.

I try to prove this is true. But I somehow fail. Here is my attempt: My process follows like this, suppose now $$p$$ is not square and I want to solve $$(q+ku)^2=1+p_1k.$$ Now I can $$q^2+2kuq+k^2u^2=1+pk+2kuq+k^2u^2=1+p_1 k.$$ I get the following Diophantine equation:

$$p+2uq+ku^2=p_1=\theta^2.$$ Now I want to solve $$u$$ in order to let $$u$$ be an integer, I must have the discriminant of a quadratic equation $$u$$ to be a perfect square. So I must have

$$4q^2-4k(p-\theta^2)=4\phi^2$$ for some integer $$\phi.$$ So know I can plug $$q^2=1+pk$$ then I can get the following Pell's equation.

$$1+k\theta^2=\phi^2.$$ Due to the theory of Pell's equation, if $$k$$ is not a perfect square, then there will be infinite solutions. But in order for $$u$$ to be an integer. I have $$u=\frac{-q \pm \phi}{k}.$$ I will need to further require that $$\phi-q$$ or $$\phi+q$$ must be divisible by $$k$$ which I am not sure how to proceed. I guess since there are infinite solutions, so it would be ok or due to some quadratic residue issue. I can not find the answer. Bu t I believe if (1) can not satisfied and then we can do for $$-q$$ then one can always make another true.

Due to some reason, I only need either (1) or (2) is true if one of them can not be satisfied. Any comments are very welcome. Thanks a lot.

• $19^2=1+10\times36$, so we can take $q=19$, $k=36$. But $(ku\pm q)^2=1+r^2k=1+36r^2=1+(6r)^2$ requires two squares to differ by $1$, which doesn't happen (except for zero and one, which no value of $u$ achieves). Oct 30, 2021 at 5:18
• Any thoughts on my comment? Oct 31, 2021 at 12:20
• Ok, thanks a lot. You beat me. Nov 1, 2021 at 13:37
• Oh, so due to the theory of Pell's equation I mentioned above, I hope when k is not a perfect square, the conjecture will be true. I will keep thinking about it. Nov 1, 2021 at 13:38