2 variable functional equation in integers

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Source: IITB Mathathon 2022 Round 2 P5

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#1 h Dec 20, 2022, 12:08 PM • 3 Y

Y by Mango247, Mango247, Mango247

Find all functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
$f(x+f(xy))=f(x)+xf(y)$ for all integers $x,y$ .

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#2 h Dec 20, 2022, 2:37 PM • 4 Y

Y by Supercali, Mango247, Mango247, Mango247

Let $P(a,b)$ be the assertion $x=a,y=b$ of the beginning functional equation.
$P(-1; -1)\Rightarrow f(f(1) - 1) = 0$ .
We examine two cases:
$\bullet$ There exists $a\neq 0$ such that $f(a) = 0$ : $P(a;1) \Rightarrow af(1)=0$
$\Rightarrow f(1) = 0\Rightarrow f(-1) = 0$ .
$P(1;y)\Rightarrow f(f(y) + 1) = f(y),\forall y\in\mathbb Z$ .
$P(-f(y) - 1; -1)\Rightarrow f(-f(y) - 1 + f(f(y) + 1)) = f(-f(y) - 1),\forall y\in\mathbb Z$
$\Rightarrow f(-f(y) - 1 + f(y)) = f(-f(y) - 1),\forall y\in\mathbb Z\Rightarrow f(-f(y) - 1) = 0,\forall y\in\mathbb Z$ .
$P(-1; f(y) + 1)\Rightarrow f(-1 + f(-f(y) - 1)) = -f(f(y) + 1),\forall y\in\mathbb Z$
$\Rightarrow f(f(y) + 1) = 0,\forall y\in\mathbb Z$ .
$P(1; y)\Rightarrow f(f(y) + 1) = f(y),\forall y\in\mathbb Z\Rightarrow f(y) = 0,\forall y\in\mathbb Z$ .
$\bullet$ $f(x)\neq 0,\forall x\neq 0$ : Then $f(1) - 1=0\Rightarrow f(1) = 1$ .
$P(1; y)\Rightarrow f(1 + f(y)) = 1 + f(y),\forall y\in\mathbb Z$ . $(I)$
Replacing $y=1$ in $(I)$ we have $f(2) = 2$ . By induction, we have $f(y) = y,\forall y\in\mathbb N^*$ .
Easily, we can prove $f(x)=x,\forall x\in\mathbb Z$ .

This post has been edited 1 time. Last edited by Hoanglahoang, Dec 20, 2022, 2:38 PM

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#3 h Apr 30, 2025, 2:34 AM

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Supercali wrote:

Find all functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
$f(x+f(xy))=f(x)+xf(y)$ for all integers $x,y$ .

Let $P(x,y)$ be the assertion $f(x+f(xy))=f(x)+xf(y)$ .
$P(-1,-1)\Rightarrow f(f(1)-1)=0$
$P(f(1)-1,1)\Rightarrow(f(1)-1)f(1)=0$

Case 1: $f(1)=0$
Since $f(f(1)-1)=0$ we get $f(-1)=0$ .
$P(1,x)\Rightarrow f(1+f(x))=f(x)$
$P(-1-f(x),-1)\Rightarrow f(-1-f(x))=0$
$P(-1,1+f(x))\Rightarrow\boxed{f(x)=0}$ which fits.

Case 2: $f(1)=1$
Since $f(f(1)-1)=0$ we get $f(0)=0$ .
$P(1,x)\Rightarrow f(f(x)+1)=f(x)+1$ , from this we can prove by induction that $f(n)=n$ for all $n\ge0$ .
Let $x\ne-1$ , then $x^2\ge0$ so $f\left(x^2\right)=x^2$ , and $x^2+x\ge0$ so $f\left(x^2+x\right)=x^2+x$ . We have:
$P(x,x)\Rightarrow(x+1)f(x)=f\left(x+f\left(x^2\right)\right)=f\left(x+x^2\right)=x+x^2$
so $f(x)=x$ . It remains to find $f(-1)$ , since we already found that $f(x)=x$ for all $x\ne-1$ .
$P(-1,2)\Rightarrow f(-1+f(-2))=f(-1)-f(2)$ and since $f(-1+f(-2))=f(-3)=-3$ and $f(2)=2$ we get $f(-1)=-1$ . So $\boxed{f(x)=x}$ for all $x$ , and we're done.

This post has been edited 2 times. Last edited by jasperE3, Apr 30, 2025, 2:36 AM

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