Recently I have been a bit distracted by math olympiad real/practice problems in the Algebra category. I will occasionally put solutions up over here

Problem 1: Find all polynomials \(P(x)\) with real coefficients such that for all real \(x\),

\[(x-1)P(x+1) = (x+2)P(x) \tag{1}\]

Solution: Inserting \(x=0\) and \(x=1\), we obtain the equalities: \(P(1)=0\) and \(-P(1) = 2P(0)\). Consequently, we obtain that \(P(0)=P(1)=0\). So let’s write

\[P(x) = (x-1)xQ(x)\]

We insert the above into (1). Crossing out common terms, we obtain

\[(x+1)Q(x+1) = (x+2)Q(x)\]

Consequently, \(x=-1\) is also a root, meaning that we can write

\[P(x) = (x-1)x(x+1)\hat{Q}(x)\]

We insert the above into (1):

\[\hat{Q}(x+1) = \hat{Q}(x)\]

and this means that \(\hat{Q}(x)\) is a constant. Consequently, \(P(x) = c x (x^2-1)\).

Problem 2: Determine all functions \(f: \mathbb{R} \rightarrow \mathbb{R}\) that satisfy

\[f(f(x)+y) = f(x)^2 + 2y f(x) + f(y) \tag{2}\]

for all real numbers \(x,y\).

Solution: Define \(g(y) = f(y) - y^2\). Observe that \(g(y + f(x)) = g(y)\). So \(g\) is periodic. We then have the following nice lemma:

Lemma The set of all periods of a function forms an additive subgroup.

Proof. Denote the set of all periods of a function \(g\) by \(P\). In the whole proof we take \(x \in \mathbb{R}\).
(Identity) Since \(g(x+0) = g(x)\), we have that \(0 \in P\).
(Closure under addition) Let \(p,q \in P\). Then \(g(x+p+q) = g((x+p)+q) = g(x+p) = g(x)\).
(Closure under inversion) Take \(p \in P\). Then we have that \(g(x-p) = g(x-p+p) = g(x)\). \(\square\)

So denote the set of all periods of \(g\) by \(P\). Then we have that \(g(y) + y^2 \in P\) and also \(g(y+p) + (y+p)^2 \in P\) with \(p \in P\). With some algebra we then obtain that \(2yp + p^2 \in P\). Now observe that setting \(y=0\), we get that \(p^2 \in P\). But then \(2yp \in P\) for any \(y\). Consequently, as long as \(P\) contains one more element than the zero element, \(P\) will be the whole real line. In conclusion, we have two scenarios: \(P = \{0\}\) then \(f=0\), and \(P = \mathbb{R}\) which implies that \(g(y)\) is constant, which gives that \(f(y) = y^2 + c\).