# The convergent of the continued fraction of $\sqrt{N}$ modulo $4$.

Consider the continued fraction of $$\sqrt{N}$$ for an integer $$N$$. Let $$d$$ be its period and $$P_n/Q_n$$ be its $$n$$-th convergent.

When $$d$$ is odd, Theorem 1 of this article that says that $$Q_{d-1}$$ is odd. However, I calculated the first few values of $$Q_{d - 1}$$ and they are all congruent to $$1 \pmod{4}$$.

Is it true that $$Q_{d-1} \equiv 1 \pmod{4}$$ for all odd values of $$d$$?

$$Q_{d-1}=K(a_1, \ldots, a_{d-1})$$ is a palindromic continuant, and since $$d=2n+1$$ is odd, it looks like this: $$Q_{d-1}=K(a_1, \ldots, a_n, a_n, \ldots,a_1)$$

From Concrete Mathematics (6.133) you have:

$$K(x_1, \ldots, x_{m+n})=K(x_1, \ldots,x_{m})\cdot K(x_{m+1}, \ldots, x_{m+n})+K(x_1, \ldots, x_{m-1})\cdot K(x_{m+2}, \ldots, x_{m+n})$$

and with $$m=n$$ and in your notations:

$$K(a_1, \ldots, a_n, a_n, \ldots,a_1)=K(a_1, \ldots,a_n)\cdot K(a_n, \ldots, a_1)+K(a_1, \ldots, a_{n-1})\cdot K(a_{n+2}, \ldots, a_1)$$

or (indeterminates can be reversed, see wiki link)

$$K(a_1, \ldots, a_n, a_n, \ldots,a_1)=K(a_1, \ldots,a_n)^2+K(a_1, \ldots, a_{n-1})^2$$

In other words, $$Q_{d-1}$$ is the sum of two squares, and since it is odd, one of the square must be odd and the other even (one of the form $$4x+1$$ and the other of the form $$4x$$), which implies $$Q_{d-1}\equiv 1 \mod(4)$$

• Neat, thanks for the solution. Regardless of what happen to the question, I still appreciate your effort. May 14 at 6:57