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f(x)+f(y)+f(xy)=2+f(x)f(y) ........................(1)
put x=y=1
3f(1)=2+f2(1)
f(1)=2 or 1
let f(1)=1
put x=2 &y=1
we get 2f(2)+f(1)=2+f(2)f(1)if we put f(1)=1 f(2)=1 thus f(x) will be constant
therefore f(1)=2
now put x=1/x & f(1)=2 in (1)
f(x)+f(1/x)=f(x)f(1/x)
thus f(x)=xn+1
f(4)=17
n=2
f(5)=26
ask if need any further clarification or rate if satisfied
That the functional equation f(x) + f(1/x) = f(x).f(1/x) has f(x) = xn+1 is not a standard result!
The equation can be written as f(xy)-1 = ( f(x) -1 ) ( f(y) - 1)
Setting g(x) = f(x) -1, we have g(xy) = g(x) g(y) for which all solutions are given by g(x) = |x|c for some constant c.
So that f(x) = |x|c+1
Here from f(4) = 17, we obtain c = 2 and hence f(5) = 26
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