The number of elements in the relation R = (x,y) : 4x^2+y^2

Question

The number of elements in the relation $R = \{(x,y) : 4x^2+y^2<52,\ x,y\in\mathbb{Z}\}$ is:

Accepted Answer

For each integer $x$, count integers $y$ with $y^2 < 52-4x^2$:

- $x=0$: $y^2<52 \Rightarrow |y|\leq7 \Rightarrow 15$ values. Contribution: $1	imes15=15$
- $x=\pm1$: $y^2<48 \Rightarrow |y|\leq6 \Rightarrow 13$ values. Contribution: $2	imes13=26$
- $x=\pm2$: $y^2<36 \Rightarrow |y|\leq5 \Rightarrow 11$ values. Contribution: $2	imes11=22$
- $x=\pm3$: $y^2<16 \Rightarrow |y|\leq3 \Rightarrow 7$ values. Contribution: $2	imes7=14$
- $x=\pm4$: $4(16)=64>52$, no solutions.

Total: $15+26+22+14 = 77$

Solutions · 1