Let $P(a,b,c,d)$ be the assertion $f(a+b) = f(a)+f(b)+f(c)+f(d)$ for $2ab=c^2+d^2$. Clearly $f(0) = 0$. Let $f(1) = n$. $P(a,a,a,a): f(2a)=4f(a)\implies f(2)=4n, f(4) = 16n$. $P(2,1,2,0): f(3) = 9n; P(4,1,2,2): f(5) = 25n$. Now suppose $f(k) = k^2n \forall k < m, m > 5$. (Note this is true for $m=6$.) We want to show $f(m) = m^2n$. If $m$ is even, we are trivially done ($f(m) = 4f(\frac{m}{2})$). Else note that $f(c_1) + f(d_1) = f(c_2) + f(d_2)$ whenever $c_1^2 + d_1^2 = c_2^2 + d_2^2 = x$ by comparing $P(x,2,2c_1,2d_1)$,$P(x,2,2c_2,2d_2)$. Since $m^2 + \frac{m-5}{2}^2 = (m-2)^2 + \frac{m+3}{2}^2$ and the other 3 values are $<m$, we are done. Hence $f(m) = m^2n$ for some fixed $n\in \mathbb{N}_0$ are the only solutions.

Wow,nice!

BAdly mistaken

Pluto1708 wrote: Wait isnt this RMM 2019 A1 also? That is the source. Read the source at the front of the page.

JustKeepRunning wrote: Pluto1708 wrote: Wait isnt this RMM 2019 A1 also? That is the source. Read the source at the front of the page. It is written RMM 2016 not 2019.

Let the functional equation be $P(a,b,c,d)$. Claim: $ f(n)=kn^2 $,where $ k=f(1) $ satisfies the condition. Proof : First,easily we get $ f(0)=0$ .$ P(n,n,n,n)$ implies $f(2n)=4f(n)$,in particular $ f(2)=4k$ $P(n^2,1,n,n)$ implies $f(n^2+1)=f(n^2) +k+2f(n) $ $P(n^2,2,2n,0)$ implies $ f(n^2+2)=f(n^2)+4k+f(2n) =f(n^2)+4k +4f(n) $ $P(n^2+1,1,n+1,n-1)$ implies $f(n^2+2)=f(n^2+1)+k+f(n+1) +f(n-1) $ Expessing only in $ f(n),f(n+1), f(n-1) $ we obtain $ f(n+1)=2f(n)-f(n-1)+2k$ Simple induction shows that $ f(n)=kn^2$ for all non-negative integers n,which concludes the proof.

We will first prove the following lemma: Lemma:For every positive integer $ n \geq 2 $ there are positive integers $ a $ and $ b $ with $ a+b=n $ for which there are nonnegative integers $ c $ and $ d $ such that $ 2ab=c^{2}+d^{2} $. Proof:From Lagrange four square theorem we have that there are nonnegative integers $ x,y,z,t $ with $ n=x^{2}+y^{2}+z^{2}+t^{2} $.If $ n $ isn't a perfect square,at least two of $ x,y,z,t $ are positive.Assume that $ x,y>0 $.Let $ a=x^{2}+z^{2} $ and $ b=y^{2}+t^{2} $.Then we can consider $ c=|xy-zt+xt+yz| $ and $ d=|xy-zt-xt-yz| $.If $ n $ is a perfect square,let $ p $ be a prime divisor of $ n $.Take $ a=pa',b=pb',c=pc',d=pd' $ where $ 2a'b'=c'^{2}+d'^{2} $ and $ a'+b'=\frac{n}{p} $(which isn't a perfect square). Now we will prove by strong induction that $ f(n)=n^{2}f(1) $. For $ a=b=c=d=0 $ we get $ f(0)=0 $. Now assume that there is $ n \geq 2 $ such that $ f(k)=k^{2}f(1) $ for any $ k < n $. From the lemma we have $ a,b,c,d<n $ such that $ 2ab = c^{2} +d^{2} $ and $ a+b=n $,which implies that $ f(n)=n^{2}f(1) $.