Let f be a function such that
f(x+ f(y)) = f(x) + y;
then f (2013) -
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Answer:
f(x+f(y))=f(x)+y
f(x)+f²(y)=f(x)+y
f²(y)=y
it's possible, when
f²=1
f=1
f(2013)=2013
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