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Question Number 147091 by aliibrahim1 last updated on 17/Jul/21

Answered by Olaf_Thorendsen last updated on 18/Jul/21

$$f(x,y+z)=f(x,y)f(x,z)$$$$y=z$$$$f(x,2y)=f(x,y)f(x,y)=f^2(x,y)$$$$\Rightarrow f(x,2^{n}y)=f^{2^n}(x,y)$$$$\Rightarrow f(x,2^n)=f^{2^n}(x,1)$$$$$$$$f(x,y+z)=f(x,y)f(x,z)$$$$f(x,y+z+t)=f(x,y)f(x,z+t)$$$$f(x,y+z+t)=f(x,y)f(x,z)f(x,t)$$$$$$$$f(5,100)=f(5,64+32+4)$$$$f(5,100)=f(5,64)f(5,32)f(5,4)$$$$f(5,100)=f(5,2^6)f(5,2^5)f(5,2^2)$$$$f(5,100)=f^{64}(5,1)f^{32}(5,1)f^4(5,1)$$$$f(5,100)=f^{100}(5,1)$$$$$$$$f(x+y,1)=f(x,1)+f(y,1)$$$$f(x+y+z+t+u,1)=f(x,1)+f(y+z+t+u,1)$$$$=f(x,1)+f(y,1)+f(z+t+u,1)$$$$=....etc$$$$=f(x,1)+f(y,1)+...+f(u,1)$$$$x=y=z=t=u=1$$$$f(5,1)=5f(1,1)$$$$$$$$f(5,100)=f^{100}(5,1)=5^{100}{f}^{100}(1,1)$$$$$$$$f(x+y,2)=f(x,2)+4f(xy,1)+f(y,2)$$$$x=y=1$$$$f(2,2)=f(1,2)+4f(1,1)+f(1,2)$$$$f^2(2,1)=f^2(1,1)+4f(1,1)+f^2(1,1)$$$$f^2(2,1)=2f^2(1,1)+4f(1,1)\left(1\right)$$$$$$$$f(x+y,1)=f(x,1)+f(y,1)$$$$x=y=1$$$$f(2,1)=2f(1,1)$$$$\left(1\right):4f^2(1,1)=2f^2(1,1)+4f(1,1)$$$$f^2(1,1)=2f(1,1)$$$$\Rightarrow f(1,1)=0\mathrm{or}2$$$$$$$$\mathrm{case}f(1,1)=2$$$$f(5,100)=5^{100}{f}^{100}(1,1)=5^{100}{2}^{100}={10}^{100}$$$$f(5,100)={10}^{100}=1\mathrm{Gogol}!$$$$$$

Commented by aliibrahim1 last updated on 18/Jul/21

$$thankyousoomuchsir$$

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