f( - f (x) - f (y))= 1 -x - y , in Zxz

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f( - f (x) - f (y))= 1 -x - y , in Zxz

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#1 h Nov 22, 2020, 1:16 PM

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Find all functions $f: Z \to Z$ that satisfy $f(-f (x) - f (y))= 1 -x - y$ for all $x, y \in Z$

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#2 h Nov 22, 2020, 1:43 PM

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Let $f(t)=0$ for some $t$
$y=t:f(-f(x))=1-x-t$
$f(-f(-f(x)-f(y)))=f(y+x-1)=1+f(x)+f(y)-t$
$y=0:f(x-1)=f(x)+f(0)-t+1$
So $f(y+x)=f(x+1)+f(y)-t+1=f(y)+f(x)-f(0) \to f(x)=ax+f(0)$
$f(-f(x)-f(y))=f( -a(x+y)-2f(0))=-a^2(x+y)-2af(0)+f(0)=1-x-y \to a^2=1, f(0)-2af(0)=1 \to f(x)=x-1$

Or more faster solution:
$f(-f(x)-f(0))=1-x \to f(x)$ is bijection
$f(-f(x+y)-f(0))=1-x-y=f(-f(x)-f(y)) \to f(x+y)+f(0)=f(x)+f(y) \to f(x)=ax+f(0)$
$1-x=f(-f(x)-f(0))=f(-ax-2f(0))=-a^2x-2af(0))+f(0) \to a^2=1, (2a-1)f(0)=-1 \to f(x)=x-1$

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themathematicalmantra

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#4 h Nov 22, 2020, 1:51 PM

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What Motivation was behind the last line?

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#5 h Nov 22, 2020, 1:58 PM • 1 Y

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This is Singapore MO 2019 P2

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#6 h Nov 23, 2020, 2:48 PM • 1 Y

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Compare $P(x, y+1)$ and $P(x+1, y)$ . RHS for both is $-x-y$ so $f(-f(x)-f(y+1))=f(-f(x+1)-f(y))$ using injectivity $f(x+1)-f(x)=f(y+1)-f(y)$ , since that $f$ is linear

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#7 h Mar 12, 2025, 6:01 AM

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Set $x=-f(a)-f(b)$ and $y=-f(c)-f(d)$ :
$f(a+b+c+d-2)=f(a)+f(b)+f(c)+f(d)+2.$ Setting $c=d=1$ , we get $f(a+b)=f(a)+f(b)+2f(1)+2$ , so if $g(x)=f(x)+2f(1)+2$ then $g$ is additive. Since $g$ is $\mathbb Z\to\mathbb Z$ as well, $f$ is linear so $f$ is linear, testing we get $\boxed{f(x)=x-1}$ .

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#8 h Mar 29, 2025, 7:22 PM

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DschinghisKhan wrote:

Compare $P(x, y+1)$ and $P(x+1, y)$ . RHS for both is $-x-y$ so $f(-f(x)-f(y+1))=f(-f(x+1)-f(y))$ using injectivity $f(x+1)-f(x)=f(y+1)-f(y)$ , since that $f$ is linear

Astonishing proof! Splendid. :clap2:

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