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 March 20th, 2016, 06:02 AM #1 Member   Joined: Aug 2015 From: Chiddingfold, Surrey Posts: 57 Thanks: 3 Math Focus: Number theory, Applied maths A surprising? discovery A SURPRISING DISCOVERY: I have only recently discovered the following equations of a certain function of a pair of two integers, which surprised me:- f(x, y) = {f(x+z, y+z) + f(x-z, y-z)}/2 ----------------1 f(x, y) = {f(x+1, y+1) + f(x-1, y-1)}/2-2 .-----------------2 Where z = any value, positive or negative. Of course the first equation is true for f(x,y) = x+/-y, and the second is true for f(x,y) = x+/-y-2, but there is another surprising function (non linear) which I will leave the reader to discover for now.
 March 20th, 2016, 10:40 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 $\displaystyle f(x,y)\neq \frac{f(x+z,y+z)+f(x-z,y-z)}{2}$
March 20th, 2016, 11:30 AM   #3
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Quote:
 Originally Posted by idontknow $\displaystyle f(x,y)\neq \frac{f(x+z,y+z)+f(x-z,y-z)}{2}$
magicterry did not say that was true for all functions! He said that there exist functions that do. He pointed out f(x,y)= x+ y and f(x,y)= x- y satisfy it and challenged people to find a third, non-linear, equation that satisfies it.

 March 20th, 2016, 11:46 AM #4 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 any f that is linear on all lines x=y+constant will do. I think that if f is differentiable one can show that its restriction to the line x=y+c has to be linear.
October 28th, 2018, 01:38 AM   #5
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Quote:
 Originally Posted by magicterry A SURPRISING DISCOVERY: I have only recently discovered the following equations of a certain function of a pair of two integers, which surprised me:- f(x, y) = {f(x+z, y+z) + f(x-z, y-z)}/2 ----------------1 f(x, y) = {f(x+1, y+1) + f(x-1, y-1)}/2-2 .-----------------2 Where z = any value, positive or negative. Of course the first equation is true for f(x,y) = x+/-y, and the second is true for f(x,y) = x+/-y-2, but there is another surprising function (non linear) which I will leave the reader to discover for now.
I've almost forgotten the other function which is simply x^2 + y^2 ( or X^2 - Y^2 + 2)

October 28th, 2018, 02:12 AM   #6
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Quote:
 Originally Posted by magicterry I've almost forgotten the other function which is simply x^2 + y^2 ( or X^2 - Y^2 + 2)
We've been waiting this whole time!

October 28th, 2018, 02:19 AM   #7
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Quote:
 Originally Posted by magicterry . . . simply x^2 + y^2
Did you mean x² - y²?

October 28th, 2018, 07:40 AM   #8
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Quote:
 Originally Posted by Joppy We've been waiting this whole time!
Correction :X^2 + Y^2 applies to equation 2 and X^2 - Y^2 applies to equation 1

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