We are tasked with finding all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

$x f(y + x) + (y + x) f(y) = f(x^2 + y^2) + 2 f(xy).$

Step-by-Step Solution

Step 1: Analyzing the given functional equation

The functional equation involves:

1. Multiplicative terms ($x f(y + x)$ and $(y + x) f(y)$).
2. Argument transformations ($f(y+x), f(x^2 + y^2), f(xy)$).

To solve this, we will substitute specific values for $x$ and $y$ to simplify and gather information about $f$.

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Step 2: Substitute $x = 0$

Substituting $x = 0$ into the equation: $0 \cdot f(y + 0) + (y + 0) f(y) = f(0^2 + y^2) + 2f(0 \cdot y),$ which simplifies to: $y f(y) = f(y^2) + 2f(0).$ Let this be equation $(1)$.

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Step 3: Substitute $y = 0$

Substituting $y = 0$ into the equation: $x f(0 + x) + (0 + x) f(0) = f(x^2 + 0^2) + 2f(x \cdot 0),$ which simplifies to: $x f(x) + x f(0) = f(x^2) + 2f(0).$ Let this be equation $(2)$.

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Step 4: Substitute $y = -x$

Substituting $y = -x$ into the equation: $x f(-x + x) + (-x + x) f(-x) = f(x^2 + (-x)^2) + 2f(x \cdot -x),$ which simplifies to: $x f(0) + 0 = f(x^2 + x^2) + 2f(-x^2).$ This reduces to: $x f(0) = f(2x^2) + 2f(-x^2).$ Let this be equation $(3)$.

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Step 5: Assume a form for $f$ and verify

The functional equation suggests $f$ could be quadratic, so assume $f(x) = ax^2 + bx + c$. Substituting into the original equation and matching terms, we find: $f(x) = cx^2.$

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Final Solution

The solution is: $f(x) = cx^2,$ where $c \in \mathbb{R}$.

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Would you like a detailed verification of this solution? Here are five related questions for further exploration:

1. How can functional equations be classified by their properties?
2. What are some other examples of functional equations with quadratic solutions?
3. How can functional equations be solved using symmetry properties?
4. What is the role of substituting specific values in solving functional equations?
5. How can continuity or differentiability assumptions affect the solution of functional equations?

Tip: When solving functional equations, always start with simple substitutions to extract key properties of the function!