Problem 953

Sol 1. The equation under consideration can be rewritten in the form $(x-5)(y+3)=-18$. Its solutions in integers correspond to the representations of the number -18 as the product of two integers.

Part 2. Some of the numbers $a_{1}, a_{2}, \ldots a_{n}$ are equal to +1, the others are equal to -1. Prove that

$$\begin{aligned} & 2 \sin \left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\cdots+\frac{a_{1} a_{2} \cdots \cdots a_{n}}{2^{n-1}}\right) \frac{\pi}{4} \\ & \quad=a_{1} \sqrt{2+a_{2} \sqrt{2+a_{3} \sqrt{2+\cdots+a_{n} \sqrt{2}}}} \end{aligned}$$

In particular, when $a_{1}=a_{2}=\cdots=a_{n}=1$, we have:

$$\begin{aligned} & 2 \sin \left(1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}\right) \frac{\pi}{4}=2 \cos \frac{\pi}{2^{n+1}}= \\ & \quad=\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}} \end{aligned}$$

Sol 2. We apply induction on $n$. For $n=1$, we get an obvious identity. The equality

$$\begin{aligned} & 2\left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\cdots+\frac{a_{1} a_{2} \cdot \ldots \cdot a_{n} a_{n+1}}{2^{n}}\right) \frac{\pi}{4}= \\ & =a_{1} \frac{\pi}{2}+a_{1}\left(a_{2}+\frac{a_{2} a_{3}}{2}++\frac{a_{2} a_{3} \cdot \ldots \cdot a_{n+1}}{2^{n-1}}\right) \frac{\pi}{4} \end{aligned}$$

shows that

$$\begin{aligned} & \cos 2\left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\cdots+\frac{a_{1} a_{2} \cdot \ldots \cdot a_{n} a_{n+1}}{2^{n}}\right) \frac{\pi}{4}= \\ & =-\sin \left(a_{2}+\frac{a_{2} a_{3}}{2}++\frac{a_{2} a_{3} \cdot \ldots \cdot a_{n+1}}{2^{n-1}}\right) \frac{\pi}{4} \end{aligned}$$

Using this formula and the identity $2 \sin \frac{\alpha}{2}= \pm \sqrt{2-2 \cos \alpha}$, we get

$$\begin{aligned} & 2 \sin \left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\cdots+\frac{a_{1} a_{2} \cdot \ldots \cdot a_{n} a_{n+1}}{2^{n}}\right) \frac{\pi}{4}= \\ & = \pm a_{1} \sqrt{2+2 \sin \left(a_{2}+\frac{a_{2} a_{3}}{2}+\cdots+\frac{a_{2} a_{3} \cdot \ldots \cdot a_{n} a_{n+1}}{2^{n-1}}\right) \frac{\pi}{4}} \end{aligned}$$

It is also not difficult to verify that the plus sign is always taken, since the sign of the number $a_{1}+$ $\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\cdots+\frac{a_{1} a_{2} \cdots a_{n} a_{n+1}}{2^{n}}$ coincides with the sign of the number $a_{1}$. Now, using the induction hypothesis, we obtain the required identity.