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I want to find all first degree functions that satisfy: $$f(x,y) = f(x+1, y-1) - 5 = f(x-2, y+1) + 8$$

I know that a first degree function looks like $$f(x,y) = ax+by+c$$

Please point me in the right direction.

Thank you in advance.

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HINT: $$f(x+1,y-1)-5=f(x-2,y+1)+8$$ means $$a(x+1)+b(y-1)+c-5=a(x-2)+b(y+1)+c+8$$ can you go further? from here we get $$3a-2b=13$$

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  • $\begingroup$ That would mean f(x,y) = ((13+2b)/3)x - ((13-3a)/2)y + c? $\endgroup$ – Stijn Hoste Jul 28 '17 at 23:45
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You have $$ \eqalign{ & \left\{ \matrix{ f(x,y) = f(x + 1,y - 1) - 5 \hfill \cr f(x,y) = f(x - 2,y + 1) + 8 \hfill \cr} \right. \cr & \left\{ \matrix{ ax + by + c = ax + a + by - b - 5 \hfill \cr ax + by + c = ax - 2a + by + b + 8 \hfill \cr} \right. \cr & \left\{ \matrix{ c = a - b - 5 \hfill \cr c = - 2a + b + 8 \hfill \cr} \right. \cr & \left\{ \matrix{ 2c = - a + 3 \hfill \cr 0 = 3a - 2b - 13 \hfill \cr} \right. \cr} $$

and since they are two equations in three unknowns you can solve for any two unknown in function of the remaining one.

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Since equations must hold for all $x,y$ we can put for example $x=0$ and $y=0$ (you culd put somthing else also). Then we get:

$$ f(0,0) = f(1,-1)-5 = f(-2,1)+8$$ thus we have

$$ c=a-b+c-5 = -2a+b+c+8$$

From $c=a-b+c-5$ we get $\boxed{a-b=5}\;\;\;\;\;\; (1)$

and

from $c=-2a+b+c +8$ we get $\boxed{-2a+b=-8} \;\;\;\;\;\; (2)$

Now if you add (boxed) equations $(1)$ and $(2)$ you get $-a=-3$ thus $a=3$ and $b=-2$. Yet we don't have any other limitation for $c$, so $c$ can be any number.

So solution of your problem is(are) function(s): $$ f(x,y)=3x-2y+c,\;\;\;\;\;\; c\in \mathbb{R}$$

You can easly check that this function(s) satisfyes starting equations.

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