1. Recall: A relation is an equivalence relation iff it is reflexive, symmetric, and transitive.

Step 1: Check $R_1$

Given $aR_1b \iff a^2+b^2=1, \qquad a,b\in \mathbb R.$

We test the three properties.

(i) Reflexive?

For reflexivity, we need $aR_1a \text{ for all } a\in \mathbb R.$ That means $a^2+a^2=1 \implies 2a^2=1 \implies a^2=\frac12.$ This is true only for $a=\pm \frac{1}{\sqrt2},$ not for all real $a$.

So $R_1$ is not reflexive.

Hence $R_1$ cannot be an equivalence relation.

(Indeed, it is symmetric since $a^2+b^2=b^2+a^2$, but that does not matter once reflexivity fails.)

Step 2: Check $R_2$

Given $(a,b)R_2(c,d) \iff a+d=b+c, \qquad (a,b),(c,d)\in \mathbb N\times \mathbb N.$

We test reflexive, symmetric, transitive.

(i) Reflexive

We need $(a,b)R_2(a,b)$ for every $(a,b)\in \mathbb N\times \mathbb N$.

Condition becomes $a+b=b+a,$ which is always true.

So $R_2$ is reflexive.

(ii) Symmetric

Suppose $(a,b)R_2(c,d).$ Then $a+d=b+c.$ Rearranging the same equality, $c+b=d+a,$ which is exactly $(c,d)R_2(a,b).$

So $R_2$ is symmetric.

(iii) Transitive

Suppose $(a,b)R_2(c,d) \quad \text{and} \quad (c,d)R_2(e,f).$ Then $a+d=b+c \quad \text{and} \quad c+f=d+e.$

From the first, $a-b=c-d.$ From the second, $c-d=e-f.$ Therefore, $a-b=e-f.$ So, $a+f=b+e,$ which means $(a,b)R_2(e,f).$

Thus $R_2$ is transitive.

Hence $R_2$ is an equivalence relation.

Step 3: Decide the correct option

• $R_1$ is not an equivalence relation.
• $R_2$ is an equivalence relation.

Therefore, only $R_2$ is an equivalence relation.

So the correct option is: $\boxed{\text{C}}$

Step 4: Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

They match.