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A function f is defined for integers m an n as given: f(mn) = f(m) x f(n) - f(m+n) +1001, where either m or n is equal to 1, and f(1) = 2. Prove that f(x) = f(x-1) + 1001. Help is appreciated :)

Guest Sep 10, 2018
 #1
avatar+2759 
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since f is symmetric in m and n we lose no generality in saying that m=1 and n is a free parameter

 

thus

 

f(n) = f(1)f(n) - f(1+n) + 1001 = 2f(n) - f(n+1) + 1001

 

f(n) = f(n+1)-1001

 

f(n-1) = f(n) - 1001

 

f(n) = f(n-1)+1001

Rom  Sep 10, 2018
 #2
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+1

Thank you so much for the help.

Guest Sep 11, 2018

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