Let \(P(x,y)\) be the statement \(f(f(x)+y) = x+f(f(y))\).

\(P(x,0)\) implies \(f(f(x)) = x + f(f(0))\). Applying this to \(P\) gives \(f(f(x)+y) = x+y+f(f(0))\).

\(P(x,f(0))\) implies \(f(f(x)+f(0)) = x+f(0)+f(f(0))\). \(P(0,f(x))\) implies \(f(f(0)+f(x)) = f(x)+f(f(0))\). Equating the two gives \(x+f(0)+f(f(0)) = f(x)+f(f(0))\), or \(f(x) = x + c\) for some fixed real number \(c\) for all real \(x\). It can be easily verified to satisfy the equation.

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TopNewestLet \(P(x,y)\) be the statement \(f(f(x)+y) = x+f(f(y))\).

\(P(x,0)\) implies \(f(f(x)) = x + f(f(0))\). Applying this to \(P\) gives \(f(f(x)+y) = x+y+f(f(0))\).

\(P(x,f(0))\) implies \(f(f(x)+f(0)) = x+f(0)+f(f(0))\). \(P(0,f(x))\) implies \(f(f(0)+f(x)) = f(x)+f(f(0))\). Equating the two gives \(x+f(0)+f(f(0)) = f(x)+f(f(0))\), or \(f(x) = x + c\) for some fixed real number \(c\) for all real \(x\). It can be easily verified to satisfy the equation.

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