Q.If $R_2 = \{(x, y) \mid x \text{ and } y \text{ are integers and } x^2 + y^2 = 64\}$ is a relation. Then find $R_2$.

Concept: Relation Arrow Diagram — list all integer pairs $(x, y)$ satisfying $x^2 + y^2 = 64$.

Step 1: Since $x$ and $y$ are integers, $x^2$ and $y^2$ are non-negative perfect squares. The only squares $\le 64$ are $0, 1, 4, 9, 16, 25, 36, 49, 64$.

Step 2: Find all pairs of these squares that sum to $64$:

• $0 + 64 = 64$ → $x = 0, y = \pm 8$ and $x = \pm 8, y = 0$
• $64 + 0 = 64$ (already covered above)
• $36 + 28$ — 28 is not a square
• $16 + 48$ — 48 not a square
• $4 + 60$ — no
• $1 + 63$ — no
• $9 + 55$ — no
• $25 + 39$ — no
• $49 + 15$ — no …

The core idea here is simple: we are looking for integer coordinates that lie exactly on a circle of radius 8 centered at the origin. The equation $x^2 + y^2 = 64$ is the circle, but since $x$ and $y$ must be integers, we are only picking lattice points on that circle.

A common mistake is to think of all real solutions — but the problem explicitly says $x$ and $y$ are integers. So we are solving a Diophantine equation.

1. List possible integer squares up to 64.

   The perfect squares less than or equal to 64 are:

   $0, 1, 4, 9, 16, 25, 36, 49, 64$.

2. Find pairs of these squares that sum to 64.

   We need $a^2 + b^2 = 64$ where $a = |x|$, $b = |y|$ are non-negative integers.

   Check systematically:

   • If $a^2 = 0$, then $b^2 = 64$ → $b = 8$.
   • If $a^2 = 1$, then $b^2 = 63$ → not a perfect square.
   • If $a^2 = 4$, then $b^2 = 60$ → no.
   • If $a^2 = 9$, then $b^2 = 55$ → no.
   • If $a^2 = 16$, then $b^2 = 48$ → no.
   • If $a^2 = 25$, then $b^2 = 39$ → no.
   • If $a^2 = 36$, then $b^2 = 28$ → no.
   • If $a^2 = 49$, then $b^2 = 15$ → no.
   • If $a^2 = 64$, then $b^2 = 0$ → $b = 0$.

   So the only non-negative integer pairs $(a, b)$ are $(0, 8)$ and $(8, 0)$. …