Question

The number of distinct real number pairs (a, b) such that a + b ∈ integers and a² + b² = 2 is/are

• A. 10
• B. 8
• C. 2
• D. 4

Answer 1

Answer: (B)

8

Answer 2

We have the circle:

$a^2+b^2=2$.

Let $(a,b)=(\sqrt{2}\cos\theta,\sqrt{2}\sin\theta)$. Then

$a+b=\sqrt{2}(\cos\theta+\sin\theta)=\sqrt{2}\cdot\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)=2\sin\left(\theta+\frac{\pi}{4}\right)$.

For $a+b$ to be an integer, let

$2\sin\left(\theta+\frac{\pi}{4}\right)=k, \quad k\in \mathbb{Z}$.

Since $\sin$ takes values in $[-1,1]$, we must have:

$-2\leq k\leq 2 \quad \Longrightarrow \quad k\in \{-2,-1,0,1,2\}$.

For each $k$:

• If $|k|=2$: $\sin\left(\theta+\frac{\pi}{4}\right)=\pm 1$ has one solution in $[0,2\pi)$.
• For $k=-1, 0, 1$: the equation $\sin\left(\theta+\frac{\pi}{4}\right)=\frac{k}{2}$ yields two solutions.

Thus, the total number of pairs is:

$1+2+2+2+1=8$.