We show $f(x)=0$ for all $x$. Denote $P(x,y)$ the assertion. We claim first that $f(xy)=f(x)+f(y)$ for one of $x,y\ge10^{100}+1$. First $P(a_1,a_2)$ gives $f(a_1a_2+c)=f(a_1)+f(a_2)$. Then $P(a_1a_2+c,a_3)$ gives $f(a_1a_2a_3+(a_3+1)c)=f(a_1)+f(a_2)+f(a_3)$. Finally $P(a_1a_2a_3+(a_3+1)c,a_4)$ gives $f(a_1a_2a_3a_4+(a_3a_4+a_4+1)c)=f(a_1)+f(a_2)+f(a_3)+f(a_4)$. By $P\left(\frac{a_1a_2a_3}{a_3+1}+c,a_3a_4+a_4\right)$ we have $f(a_1a_2a_3a_4+(a_3a_4+a_4+1)c)=f\left(\frac{a_1a_2a_3}{a_3+1}+c\right)+f(a_3a_4+a_4)$ so setting $a_2=a_3+1$ we get $f(a_1)+f(a_3+1)+f(a_3)+f(a_4)=f(a_1a_3+c)+f(a_3a_4+a_4)=f(a_1)+f(a_3)+f(a_3a_4+a_4)$. If $a_3+1=x,a_4=y$ we get our claim. Next we show by strong induction that $f(xy+n)=f(x)+f(y)$ for sufficiently large $y$. The base case $n=0$ is proven above. For the inductive step assume $f(xy)=f(xy+1)=\dots=f(xy+n)$. First we have $P(x,z)$ implies $f(xz+c)=f(x)+f(z)$ and $P(xz+c,y)$ implies $f(xyz+c(y+1))=f(x)+f(y)+f(z)$. Next from the inductive hypothesis we have $f(xyz+(n+1)z+i)=f(xy+n+1)+f(z)$ for $0\le i\le n$ since when $y$ is large enough so is $xy+n+1$. Then we can set $z=\frac{c(y+1)-i}{n+1}$ such that $z$ is an integer and is greater than $10^{100}$, which is possible for large $y$. This implies $f(xy+n+1)=f(x)+f(y)$ for large $y$, completing the induction. Eventually we get $n>10^{100}$ so setting $x=n$ we have $f(n)+f(y)=f(ny)=f(ny+n)=f(n)+f(y+1)$ so $f(y)$ is eventually constant, and $P(y,y)$ gives this constant is zero. Now induct downward, taking $P(y-1,y)$ to get $f$ is constantly zero.

We start with the functional equation: \[ f(m) + f(n) + f(k) = f(mn + c) + f(k) = f(mnk + kc + c) \] Substituting \( n = kc \): \[ f(m) + f(kc) + f(k) = f(mk^2c + kc + c) = f(kc) + f(mk+1) \] Thus, we obtain: \[ f(m) + f(k) = f(mk+1) \] Similarly, \[ f(mnk+1) = f(n) + f(mk) = f(mn) + f(k) \] This implies that \( f(mn) - f(n) \) is a function only of \( m \), so we define: \[ f(mn) - f(n) = g(m) \] Similarly, \[ f(mn) - f(m) = g(n) \] From this, we deduce: \[ f(n) - g(n) = \text{const} = d. \] Thus, \[ f(nm) = g(nm) + d = g(n) + g(m) + d = f(n) + f(m) - d = f(nm+1) - d. \] Now, take \( n > 10^{100} \) and a sufficiently large \( k \) such that both \( k \) and \( k-1 \) are expressible as the product of two numbers greater than \( 10^{100} \). Then: \[ 2f(k) = f(k-1) + f(k+1) = f(k^2-1) + d = f(k^2) = 2f(k) - d \] Thus, we conclude: \[ d = 0. \] Now, using: \[ f(n+1) + f(n-1) = f(n^2 - 1) = f(n^2) = 2f(n), \] we see that \( f \) follows an arithmetic progression. Since sometimes the difference is 0, \( f \) must be a constant function. The only constant that works is: \[ f(n) = 0. \]

Answer is $f\equiv 0$. Lemma: If a number can be written as a product of two numbers from $S$, then call that number special. We can find an arithmetic progression with length $l\in Z^+$ and common difference $c$ where each term is special (also we can provide a number to divide the first term of this sequence). Proof: Let $\{a,a+c,\dots,a+(l-1)c\}$ be the arithmetic progression. Pick $10^{100}<p_1<p_2<\dots <p_l<q_1<\dots<q_l$ primes. Let $a\equiv -ic(mod \ p_{i+1}q_{i+1})$ which exists by Chinese Remainder Theorem. This sequence satisfies the conditions. Claim: $f(xy)=f(x)+f(y)$. Proof: We have $f(y)+f(xz)=f(xyz+c)=f(x)+f(yz)$ hence $f(xz)-f(x)$ is determined according to only $z$. Define $g(z)=f(xz)-f(x)$. We observe that $f(z)+g(x)=f(xz)=f(x)+g(z)$ thus, $f(x)-g(x)$ is constant. Set $g(x)=f(x)+k$. So $f(xy)=f(x)+f(y)+k$. Note that $f(xy+c)=f(x)+f(y)=f(xy)-k$. If $x$ is special, then $f(x+c)=f(x)-k$. By Lemma, we can find an arithmetic sequence with length $x+1$ such that each term is special. Thus, $f(x)+f(y)+f(z)=f(x)+f(yz+c)=f(xyz+cx)=f(xyz)-xk=f(x)+f(y)+f(z)+2k-xk$ hence $k(x-2)=0$. Since $x>2$, we have $k=0$ which proves the claim. Claim: $c$ is a period of $f$. Proof: By the proven claim, we get that $f(x+c)=f(x)$ if $x$ is a special number. Pick $xy\equiv -ic(mod \ p_{i+1}q_{i+1})$ for sufficiently large distinct primes. Each of $\{xy,xy+c,\dots,xy+cx\}$ are special thus, $f(x)+f(y+c)=f(xy+cx)=f(xy)=f(x)+f(y)$ which implies $f(y)=f(y+c)$ for any $y\in S$. Claim: $f\equiv 0$. Proof: We observe that $f(cx)+f(y)=f(cxy)=f(cxy+cx)=f(cx)+f(y+1)$ yields $f(y)=f(y+1)$ so $f$ is constant. Since $f(xy)=f(x)+f(y)$, we get $f\equiv 0$ as desired.$\blacksquare$

The answer is $\boxed{f\equiv 0}$ only, which obviously works. This problem and P1 are similar in that they both feel hopeless until you solve it. Let $L=10^{100}$. If $r>1$ is an integer, then \[f(rx)+f(y)=f(rxy+c)=f(x)+f(ry)\Longrightarrow f(rx)-f(x)=d_r\]for some constant $d_r$. Now it is not hard to show that $d_m+d_n=d_{mn}$. So now define $\alpha=f(L)-d_{L}$. Then note that \[f(x)=f(xL)-d_L=f(L)+d_x-d_L=\alpha+d_x.\]So we get \[f(p_1^{e_1}p_2^{e_2}\dots)=\alpha+d_{p_1}e_1+d_{p_2}e_2+\dots\]Now note that \[f(x(cy)+c)=f(x)+f(cy)\Longrightarrow f(xy+1)=f(x)+f(y).\]Now we get \[f(xy+1)=f(x)+f(y)=f(xy)+\alpha.\]Let $T$ be the set of numbers that can be expressed as $xy$ where $x,y\in S$. Note that if $t\in T$, then $f(t+1)=f(t)+\alpha$. Then the natural density of $T$ is $1$, so we can find some $u$ such that $u\in T$ and $2u+1\in T$. Now let $u=xy$, then \[f(2xy+1)=f(2xy)+\alpha=d_2+f(xy)+\alpha=d_2+f(xy+1)=f(2xy+2)=f(2xy+1)+\alpha,\]so $\alpha=0$, so $f(xy)=f(x)f(y)$. Now if $x\ge L+1$, we get \[f(x)+f(x)=f(x^2+1)=f(x^2)=f(x-1)+f(x+1),\]so $f$ is linear, say $f(x)=ax+b$. Then $(ax+b)(ay+b)=axy+b$, so $b=0$ and $a\in\{0,1\}$. But $a=1$ doesn't work, so we are done. $\blacksquare$

The answer is $f(x)=0,$ which obviously works. First, notice that, if $x_{1}y_{1}=x_{2}y_{2},$ then $f(x_{1})+f(y_{1}) = f(x_{2})+f(y_{2}).$ So, substituting in $x, abx, ax, bx$ for $x_{1}, y_{1}, x_{2}, y_{2},$ we get $f(abx)-f(bx) = f(ax)-f(x),$ and thus $f(am)-f(m) = f(an)-f(n)$ if $m \mid n.$ So, $f(ax)-f(x) = f(axy)-f(xy) = f(ay)-f(y),$ where $x$ and $y$ are arbitrary. Thus, we define $g(a) = f(ax)-f(x),$ where $x$ can be any integer in $S.$ Notice that applying $P(cx,y)$ gives $f(cxy+c) = f(cx)+f(y),$ and so subtracting both sides by $g(c),$ we get $f(xy+1) = f(x)+f(y).$ Claim:$g(ab) = g(a)+g(b).$ Proof: Notice that $g(ab)-g(a) = (f(abx)-f(x))-(f(ax)-f(x)) = f(abx)-f(ax) = g(b),$ so $g(b)+g(a)=g(ab).$ Claim: $g(x) = f(x)-a,$ for some $a.$ Proof: We have $f(y)+g(x) = f(y)+(f(xy)-f(y)) = f(xy),$ and $f(x) + g(y) = f(xy),$ so $f(x)-f(y) = g(x)-g(y).$ Thus, $f(x)-g(x) = f(y)-g(y),$ as desired. $\blacksquare$ So, $-a = g(x)-f(x) = f(xy)-f(y)-f(x),$ so $f(xy) = -a+f(x)+f(y),$ and thus $f(xy) = -a+f(xy+c).$ Claim: There exist three consecutive positive integers $b-1,b,b+1$ that all can be expressed in the form $xy,$ where $x,y\in S.$ Proof:Take $x = 10^{100}+1,$ let $y = x+1,$ and let $z = x+2.$ Notice that these are all in $S,$ and are pairwise relatively prime. Then, $x+Nxyz, y+Nxyz,z+Nxyz $ are consecutive, and are multiples of $x,y,z.$ Taking $N$ large enough guarantees that dividing these numbers by $x,y,$ or $z$ will keep them in $S.$ So, we have proven the claim. $\blacksquare$ Now, notice that $$f(b-1)+f(b+1) = f(b^{2}) = f(b)+f(b)-a,$$and $b-1,b,b+1$ can be written as $xy,$ for $x,y \in S,$ so $f(b-1) = f(b)-a,$ and $f(b+1) = f(b)+a,$ so $f(b-1)+f(b+1) = 2f(b),$ and thus $a=0.$ So, $f(xy+1) = f(xy) = f(x)+f(y).$ Now, $f(b^{2}) = f(b-1)+f(b+1) = f(b)+f(b),$ so $f(b)-f(b-1)=f(b+1)-f(b)$, and thus $f(b+1)-f(b)$ is constant. When $b = xy,$ this is equal to zero, so $f(b+1)=f(b)$ for all $b,$ and thus $f$ is constant. Applying the functional equation to any $x,y$ clearly gives $f(x)=0,$ and so we are done.

Notice that $f(x)+f(x+1)+f(y) = f(xy+c) + f(x+1) = f(x(x+1)y + cx + 2c)$ and $f(x)+f(x+1)+f(y) = f((x+1)y+c) + f(x) = f(x(x+1)y+cx+c)$. Thus taking $x$ great enough and divisible by $c$, we get the existence of $u \in S$ such that $f(cu) = f(cu+c) \iff f(u-1)+f(c) = f(u) +f(c) \iff f(u-1) = f(u)$. Now, write $ f(u-1)+f(c) + f(u+1)= f(uc) + f(u+1)= f(u(u+1)c + c) = f(c(u+1))+f(u) = f(u) + f(u) +f(c) \implies f(u+1)=f(u)$. Consequently, we have $f(u-1)=f(u)=f(u+1) = \ldots$, which means $f$ is eventually constant. Finally, if there exists $b \in S$ such that $|f(b)| > 0$, then $|f(b^2+c) | =2|f(b)|$. As a consequence, we get the existence of arbitrarily large integers $n$ such that $f(n)$ can be as large as we want, which is a contradiction with the previous result, and thus implies $|f(b)| = 0$ for every $b \in S$. It's obvious that this solution works.

We will show that $f(x)=0$ for all $x\in S$. Let $M\doteqdot 10^{100}$. Claim. $f(xy+1)=f(x)+f(y)$ for all $x,y\in M$. Proof. Let $P(x,y)$ be the assertion $f(xy+c)=f(x)+f(y)$ for all $x,y\ge M$. $$P(xy+c,z)\implies f(xyz+zc+c)=f(xy+c)+f(z)=f(x)+f(y)+f(z).$$Setting $y=zc$ implies that $$f(x)+f(zc)+f(z)=f(xz^2c+zc+c)=f(zc)+f(xz+1).$$Hence $f(xy+1)=f(x)+f(y)$ for all $x,y\in S$. $\square$ Now let $n\ge M+1$. The claim above implies that $$f(n^4+1)=2f(n^2)=2[f(n-1)+f(n+1)]\qquad(1)$$Moreover, $$f(n^4+1)=f(n)+f(n^3)=f(n)+f(n-1)+f(n^2+n+1)=f(n-1)+2f(n)+f(n+1)\qquad(2)$$From $(1)$ and $(2)$ we conclude that $$f(n-1)+f(n+1)=2f(n)\qquad(3)$$This proves that $f(n)-f(n-1)$ is constant. Call this constant $d$. From $(3)$ we conclude that $$f(n^2)=f(n-1)+f(n+1)=2f(n)=f(n^2+1).$$Hence $d=0$ so $f$ is constant. Call this constant $c$. The identity $f(M^2+1)=2f(M)$ implies that $c=2c$ so $c=0$, as desired.

Fakesolve

First of all let $P(x,y)$ denote the assertion. Note that by using $P(ax, y)$ and $P(ay, x)$ we can say that $f(ax)-f(x)=g(a)$ is constant(well actually it's depended on $a$). Now put in $P(x, yc)$ we get $f(x)+f(cy)=f(c(xy+1))=f(xy+1)+g(c)$ and $f(xy+1)=f(x)+f(cy)-g(c)=f(x)+f(y)$. So $f(x)+f(y)=f(xy+1)$. Now put $P(x, (x+1)z+1)$ for some $z \ge 10^{100}$ we get $f((x+1)(xz+1))=f(x)+f((x+1)z+1)$ which means $f(xz+1)+g(x+1)=f(x)+f((x+1)z+1)$ $\implies$ $g(x+1)=f(x+1)$. Now because $f(x)+g(a)=f(a)+g(x)$ we get $f(x)=g(x)$ for all $x \ge 10^{100}$ which means $f(x)+f(y)=f(xy)$. Now by putting $x+1, x+1$ and $x, x+2$ we get $f$ is an arithmetic sequence, then we get $f(x)=0$ for all $x$ obviously.

This was proposed by Will Steinberg, United Kingdom. Observe that if $x_1,y_1,x_2,y_2\in S$ with $x_1y_1=x_2y_2$ then $$ f{\left(x_1\right)}+f{\left(y_1\right)}=f{\left(x_2\right)}+f{\left(y_2\right)}. \quad \quad \left(\blacktriangle\right) $$This tells us that for $u,v,w\in S$ we have $$f{\left(uv\right)}+f{\left(w\right)}=f{\left(u\right)}+f{\left(vw\right)}$$$$\Rightarrow f{\left(uv\right)}-f{\left(u\right)}-f{\left(v\right)}=f{\left(vw\right)}-f{\left(w\right)}-f{\left(v\right)}.$$Notice the $\mathrm{RHS}$ is independent of $u$ so the same must be true of the $\mathrm{LHS}$. By replicating the argument with $u$ and $v$ switched, we also see the $\mathrm{LHS}$ is independent of $v$ so in fact $$f{\left(uv\right)}-f{\left(u\right)}-f{\left(v\right)}=k \quad \text{for some constant} \quad k \in \mathbb{Z}. \quad \quad \left(\blacksquare\right)$$Using $\left(\blacktriangle\right)$ again we have, for $y,z \in S$ $$f{\left(cz\right)}+f{\left(y\right)}=f{\left(z\right)}+f{\left(cy\right)}$$$$\Rightarrow f{\left(cy\right)}-f{\left(y\right)}=l \quad \text{for some constant} \quad l \in \mathbb{Z}. \quad \quad \left(\bigstar\right)$$Setting $x=cz$ in the original functional equation for $z \in S$ shows $$f{\left(c(yz+1)\right)} \stackrel{\left(\bigstar\right)}{=} f{\left(yz+1\right)}+l=f(cz)+f(y)\stackrel{\left(\bigstar\right)}{=} f(z)+f(y)+l$$$$\Rightarrow f(yz+1)=f(y)+f(z).$$Let $x \in S$ and set $y=x$, $z=x+2$ in the above to get $$f{\left((x+1)^2\right)} \stackrel{\left(\blacksquare\right)}{=} 2f(x+1)+k=f(x)+f(x+2)$$$$\Rightarrow f(x)+f(x+2)-2f(x+1)=-k=\text{constant}$$which forces $f$ to be a quadratic. By setting $x=y$ in the original functional equation and considering the degree of both sides, we see $f$ must be in fact be constant. The only constant function that satisfies the condition is $\boxed{f \equiv 0}$.

Different from others, I guess. We claim that $ f \equiv 0 $. Define $ \Delta_x = f(x^2) - f(x) $. From the given functional equation, we derive the following: 1. $ f(x^n) = f(x) + (n-1)\Delta_x $. 2. $ f(xy) - f(y) = \Delta_x $. 3. Setting $ \varepsilon = f(x) - \Delta_x $, we find that $ \varepsilon $ is constant for all $ x \in S $. 4. The functional equation simplifies to $ f(x) + f(y) - f(xy) = \varepsilon $. From \[ f(xy+c) - f(xy) = f(x) + f(y) - (f(x) + f(y) - \varepsilon) = \varepsilon, \]induction gives \[ f(xy + Nc) - f(xy) = N\varepsilon. \]Setting \( N = y \), we obtain \[ y\varepsilon = f((x+c)y)-f(xy) = f(x+c) - f(x). \]Since $ y $ is arbitrarily large, it follows that $ \varepsilon = 0 $, so $ f(x+c) = f(x) $, meaning $ f $ has period $ c $. \[ f(x) + f(x+1) = f(x^2 + x + c)=f(x^2 + 2cx + c^2 + c) = f((x+c)^2 + c) = 2f(x+c) = 2f(x), \]which implies $ f(x+1) = f(x) $, so $ f $ is constant on $ S $. Substituting back, we conclude that $ f \equiv 0 $. I love this problem though, literally no number is substituted.