Given that $R$ is a ring such that $(a+b)^2 = a^2 + b^2$ for all $a, b \in R$, we need to show that $2a = 0$ for all $a \in R$.

First, let's expand the left-hand side of the given equation using the distributive property:

$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2$

According to the given condition, this should equal the right-hand side:

$(a + b)^2 = a^2 + b^2$

Equating the two expressions, we get:

$a^2 + ab + ba + b^2 = a^2 + b^2$

Now, subtract $a^2 + b^2$ from both sides:

$ab + ba = 0$

This implies that for any $a, b \in R$:

$ab + ba = 0$

Next, let's consider $b = a$:

$a^2 + a^2 = 0$

Simplifying this, we get:

$2a^2 = 0$

In a ring, $2a^2$ can be written as $2 \cdot a^2$, where $2$ is the integer considered as an element of the ring $R$. This shows that $2 \cdot a^2 = 0$. However, since this must hold for any $a \in R$, we generalize it to:

$2a = 0$

Therefore, we have shown that $2a = 0$ for any $a \in R$.

Do you want more details or have any questions?

Here are 5 more questions for you to consider:

1. What are some examples of rings where $2a = 0$?
2. How does the property $(a+b)^2 = a^2 + b^2$ relate to commutativity in a ring?
3. Can we extend this property to fields, and what would be the implications?
4. What are zero divisors in ring theory, and how do they relate to this problem?
5. How does the characteristic of a ring affect properties like $2a = 0$?

Tip: When solving problems in abstract algebra, try to use general properties and identities of the structures involved to simplify and understand the problem better.