Any solutions

abbosjon2002 wrote: Find all functions from positive integers to itself such that $f(a+b)=f(a)+f(b)+f(c)+f(d)$ for all $c^2+d^2=2ab$ Let $P(x,y,z,t)$ be the assertion $f(x+y)=f(x)+f(y)+f(z)+f(t)$ true $\forall x,y,z,t\in\mathbb Z_{>0}$ such that $2xy=z^2+t^2$ Let $a=f(1)$ and $b=f(3)$ $P(x,x,x,x)$ $\implies$ $f(2x)=4f(x)$ So $f(2)=4a$, $f(4)=16a$, $f(6)=4b$ and $f(8)=64a$ $P(4,1,2,2)$ $\implies$ $f(5)=25a$ $P(9,1,3,3)$ $\implies$ $f(9)=99a-2b$ Let positive integer $x\ge 3$ $P(10x^2+10,1,2x+4, 4x-2)$ $\implies$ $f(10x^2+11)=f(10x^2+10)+f(1)+f(2x+4)+f(4x-2)$ $P(10x^2+10,1,2x-4, 4x+2)$ $\implies$ $f(10x^2+11)=f(10x^2+10)+f(1)+f(2x-4)+f(4x+2)$ Subtracting, we get $f(2x+4)+f(4x-2)=f(2x-4)+f(4x+2)$, which is : Assertion $Q(x)$ : $f(2x+1)=f(2x-1)+f(x+2)-f(x-2$ true $\forall$ integer $x\ge 3$ $Q(3)$ $\implies$ $f(7)=49a$ $Q(4)$ $\implies$ $f(9)=4b+45a$ and so $4b+45a=99a-2b$ and so $b=9a$ and $f(9)=81a$ So $f(n)=n^2a$ $\forall n\in\{1,2,3,4,5,6,7,8,9\}$ And $Q(x)$ allows simple induction for all odd $n$ (and even $n$ are obtained from odd ones) And so $\boxed{f(x)=ax^2\quad\forall x\in\mathbb Z_{>0}}$ which indeed is a solution, whatever is $a\in\mathbb Z_{>0}$