Cool functions

Find all functions f:RRf:\mathbb{R} \rightarrow \mathbb{R} such that

f(f(x)+y)=x+f(f(y))f(f(x)+y)=x+f(f(y))

for all real numbers xx and yy.

#Algebra #Sharky

Note by Sharky Kesa
5 years, 7 months ago

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Comments

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

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

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

Ivan Koswara - 5 years, 7 months ago
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