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Consider the function [math]f:\mathbb{Z}\to\mathbb{Z}[/math] defined by [math]f(n)=n(n-1)(n+1).[/math]If we evaluate [math]f(r+1)-f(r)[/math], then we obtain:[math](r+1)r(r+2)-r(r-1)(r+1)=r(r+1)((r+2)-(r-1))=3r(r+1).[/math]Hence[math]f(r+1)-f(r)=3r(r+1).[/math]Thus[math](f(1)-f(0))+(f(2)-f(1))+(f(3)-f(2))+\cdots+(f(n+1)-f(n))[/math][math]=3(0)(1)+3(1)(2)+3(2)(3)+\cdots+3n(n+1).[/math][math]f(n+1)-f(0)=3(1\times 2+2\times 3+\cdots+n(n+1)).[/math][math](n+1)(n)(n+2)=3(1\times 2+2\times 3+\cdots+n(n+1)).[/math][math]1\times 2+2\times 3+\cdots+n(n+1)=\dfrac{n(n+1)(n+2)}{3}.[/math]

I am not so experienced in these questions but I will give it a tryFirst of all if we plug in [math][/math][math]\ [/math][math][/math][math]\ x=y=0\ [/math][math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math][/math]we get[math]f(0^2–0^2)=f^2(0)-f^2(0)\iff \boxed {f(0)=0}[/math]Then for plug in [math][/math][math]\ [/math][math][/math][math]\ y=0\ [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math][/math]and[math]f(x^2–0^2)=f^2(x)-f^2(0)\iff \boxed {f(x^2)=f^2(x)}[/math]We can also notice that[math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math]f[/math][math] [/math][math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math][/math]is odd plugging in [math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ x=0 [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math][/math]because[math]f(0^2–y^2)=f^2(0)-f^2(y)\iff [/math][math][/math][math][/math][math]f(–y^2)=-f^2(y)\iff[/math][math]f(–y^2)=-f(y^2)\iff[/math][math]f(–t)=-f(t)\implies [/math][math][/math][math] [/math][math][/math][math] [/math][math]f[/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math][/math]is odd (actually we don’t need this in order to answer the question but it is good to know anyway)Now lets go for [math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] f(1) [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math][/math]plugging in[math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] x=1 [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] and [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math] y=0 [/math][math][/math][math][/math][math]f(1^2–0^2)=f^2(1)-f^2(0)\iff [/math][math][/math][math][/math][math]f(1)=f^2(1)\iff [/math][math][/math][math][/math][math]f^2(1)-f(1)=0\iff [/math][math][/math][math][/math][math]f(1)[f(1)-1]=0\iff [/math][math][/math][math][/math][math]f(1)=0 [/math][math][/math][math]\ \text{ or} [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] f(1)=1[/math]but[math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] f(1)\neq 0[/math](Look at this beautiful explanation of [math][/math][math]\ [/math][math][/math][math]\ f(1)\neq 0\ [/math][math][/math][math]\ [/math][math][/math][math][/math] which made me adopt it and delete my explanation and make a link to this [1][1][1][1]!)Using David Shaffer's explanation[math]\boxed {x=\sqrt{z},y=1 \implies [/math][math][/math][math]\\ f(z-1)=f^2(\sqrt{z})-f^2(1)=f((\sqrt z)^2)-f(1^2)=f(z)-f(1)\\ [/math][math][/math][math]\ \text {So}\ [/math][math][/math][math]\ [/math][math][/math][math] f(z)=f(z-1)+f(1)\\ \text{Summing from} [/math][math][/math][math]\ [/math][math][/math][math]\ z=1 [/math][math][/math][math]\ [/math][math][/math][math]\ \text{to} [/math][math][/math][math]\ [/math][math][/math][math]\ z=100 \text{ [/math][math][/math][math] and cancelling the common terms:}\\f(100)=f(0)+100f(1)\\ f(100)=100f(1)\\ \text{We are given} [/math][math][/math][math]\ [/math][math][/math][math]\ f(100)\ne 0 [/math][math][/math][math]\ [/math][math][/math][math]\ \text {so} [/math][math][/math][math]\ [/math][math][/math][math]\ f(1) \ne 0}[/math]so[math]\boxed [/math][math][/math][math]{ f(1)=1 [/math][math][/math][math]}[/math]now lets plug in[math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] x=\sqrt{t+1} [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math][/math]and [math][/math][math]\ [/math][math][/math][math]\ y =\sqrt{ t [/math][math][/math][math]}[/math][math]f((\sqrt{t+1})^2 -(\sqrt{ t })^2)=f^2(\sqrt{t+1})-f^2(\sqrt{t})\iff [/math][math][/math][math][/math][math]f(t+1 - t )=f((\sqrt{t+1})^2)-f((\sqrt{t})^2)\iff [/math][math][/math][math][/math][math]f(1 [/math][math][/math][math] )=f(t+1)-f(t)\iff[/math][math][/math][math] f(t+1)-f(t)=1 [/math][math][/math][math] \iff[/math][math]\boxed {f(t+1)=f(t)+1} [/math][math][/math][math] [/math][math][/math][math][/math]but then since we know now that [math][/math][math]\ [/math][math][/math][math]\ f(0)=0[/math]we get[math]f(1)=f(0+1)=f(0)+1=0+1=1[/math][math]f(2)=f(1+1)=f(1)+1=1+1=2[/math]and generallyif[math][/math][math] [/math][math][/math][math] [/math][math][/math][math]\ [/math][math][/math][math]\ f(n)=n [/math][math][/math][math]\ [/math][math][/math][math]\ [/math][math][/math][math][/math]then[math]f(n+1)=f(n)+1=n+1 [/math][math][/math][math][/math]So[math]\forall [/math][math][/math][math]\ x \in \N, [/math][math][/math][math] f(x)=x [/math][math][/math][math] [/math][math][/math][math][/math]which is what we need to know to figure out that[math][/math][math] [/math][math][/math][math] \boxed [/math][math][/math][math]{ [/math][math][/math][math]\ [/math][math][/math][math]\ f(2019)=2019\ [/math][math][/math][math]\ [/math][math][/math][math] [/math][math][/math][math]}[/math]Footnotes[1] David Shaffer's answer to Suppose a function f:R→R satisfies f(100) ≠0, f (x^2-y^2) =f(x) ^2-f(y) ^2 for all x,y. What is the value of f (2019)?[1] David Shaffer's answer to Suppose a function f:R→R satisfies f(100) ≠0, f (x^2-y^2) =f(x) ^2-f(y) ^2 for all x,y. What is the value of f (2019)?[1] David Shaffer's answer to Suppose a function f:R→R satisfies f(100) ≠0, f (x^2-y^2) =f(x) ^2-f(y) ^2 for all x,y. What is the value of f (2019)?[1] David Shaffer's answer to Suppose a function f:R→R satisfies f(100) ≠0, f (x^2-y^2) =f(x) ^2-f(y) ^2 for all x,y. What is the value of f (2019)?

This question is a bit old, but I noticed all the other solutions have flaws in them, so I felt obligated to solve it myself. I encourage you to try and find the mistake in the other proofs. It’s a mistake made by at least 95% of people when they first get into functional equations, so it you can identify it, then you’re on the right track!If however you’re feeling lazy, I can convince you that their solutions is wrong:Proof 1: This problem is P5 from the International Mathematical Olympiad 2015, so it is extremely unlikely that it can be solved in a few lines.Convinced? No? Okay how about this:Proof 2: [math]f(x)=2-x[/math] is a solution ( you can check) but none of the other answers included it, so they must have a flaw.Now that we got that out of the way, let’s do some substitutions, and what better place to start than [math]0[/math].for [math]y=0[/math] :[math]f(x+f(x))=x+f(x) - f(0) \tag{1}[/math]for [math]x=0[/math]:[math]f(f(y))+f(0)=f(y)+yf(0) \tag{2}[/math]for [math]y=1[/math] :[math]f(x+f(x+1)) =x+f(x+1)\tag{3}[/math]Now substitute [math]y=x+f(x+1)[/math] in [math](2)[/math] (keeping in mind that [math]y=f(y)=f(f(y))[/math]) we get :[math]f(0)=yf(0)\tag*{}[/math]This could mean two things:If [math]f(0) \neq 0[/math] then [math]y=1 \Leftrightarrow x+f(x+1)=1 \Leftrightarrow f(x+1)=1-x[/math] setting [math]z =x+1[/math] gives us the possible solution [math]f(z)=2-z[/math].If [math]f(0)=0[/math], then we [math](1)[/math] and [math](2)[/math] can be simplified so we get:[math]f(x+f(x))=x+f(x)\tag{4}[/math]and :[math]f((y))=f(y)\tag{5}[/math]You should be convinced that the only solution in this case is the identity. How do we show that? Well, since we know[math][/math][math] f(f(y))=f(y)[/math] it suffices to show that [math]f[/math][math][/math] is surjective or injective. Judging by the equation we have, proving surjectivity seems unlikely but injectivity seems promising…*[math]30[/math] minutes later*%$^# this. Okay maybe it wasn’t that promising. We need to go another route.What do we do? Well, we know that [math]f(0)=0[/math]. so we can make use of it by setting [math]y=-x[/math] in our original equation :[math]f(x+f(0)) [/math][math][/math][math]+ f(-x^2)=x+f(0)-xf(x) \tag*{}[/math][math]\Leftrightarrow f(-x^2)=x -(x+1)f(x)\tag{6}[/math]Interesting, I can use this to figure out [math]f(-1)[/math] by plugging in [math]x=-1[/math], which would yield [math]f(-1)=-1[/math] but then I can set [math]x=1[/math] (in the same equation) and get that [math]f(1)=1[/math]. Good Stuff. More importantly though, LHS of [math](6)[/math] does not change if I replace [math]x[/math] with [math]-x[/math], so same should apply for RHS. Concretely this means that:[math]x-(x+1)f(x)=f(-x^2)=-x-(-x+1)f(-x) \tag{7}[/math]Hmmm.. We know that [math]f[/math][math][/math] is probably the identity, so it should be an odd function, if I can prove that, and replace [math]f(-x)[/math] with [math]-f(x)[/math] in [math](7)[/math] then I would get an equation involving only [math]f(x)[/math] and [math]x[/math], and would be able to determine the function.Objective: Prove that [math]f[/math][math][/math] is an odd function.To do this, I want to make a substitution that would make [math]x[/math] and [math]-x[/math] appear in my equation, and since I already tried [math]y=-x[/math], let’s try something else now, how about [math]y=-1 [/math][math][/math][math][/math]:[math]f(x+f(x-1))+f(-x)=x+f(x-1) -f(x)\tag*{}[/math]Now if only I had [math]f(x+f(x-1))=x+f(x-1)[/math] , then we’d be done! Can we show that? i.e. can we prove that [math]x+f(x-1)[/math] is a fixed point? This seems feasible for the following reason: if I let [math]S[/math] be the set of points fixed by [math]f[/math][math][/math] then we know that [math]-1,0,1, f(x),x+f(x),x+f(x+1)[/math] are all elements of [math]S,[/math] so we already have some idea about the structure of the set.Turns out, all we need is this one magical substitution to finish up the problem:Let [math]x=1[/math] and [math]y=z-2 [/math][math][/math][math]+ f(z-1)=z-2 [/math][math][/math][math]+ f((z-2)+1)[/math] (and note that by [math](3)[/math], [math]y\in S[/math]) We get the following:[math]f(1+f(y+1)) [/math][math][/math][math]+ f(y)=1+f(y+1)+y\tag*{} [/math][math][/math][math][/math][math]\Leftrightarrow f(1+f(y+1))=1+f(y+1) \tag*{}[/math]so [math]1+f(y+1)[/math] is a fixed point, replacing [math]y[/math] by its value we get:[math]1+f(y+1)=1+f(z-1+f(z-1))= z+f(z-1)\tag*{}[/math]This is precisely want we wanted to show! So we conclude that [math]f[/math][math][/math] is odd.Replacing [math]f(-x)[/math] by [math]-f(x)[/math] in [math](7)[/math] we get :[math]x-(x+1)f(x)=-x-(x-1)f(x) \Leftrightarrow f(x)=x [/math][math][/math][math][/math]Finally, we can easily verify that [math]f(x)=x[/math] and [math]f(x)=2-x[/math] indeed satisfy the given equation.[math][/math][math] \blacksquare[/math]