khashi70 wrote: Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have : $ P(x^{2},y^{2}) = P(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2})$ The equation may be written $ P(\frac {x^2 + y^2}{2} + \frac {x^2 - y^2}{2},$ $ \frac {x^2 + y^2}{2} - \frac {x^2 - y^2}{2})$ $ = P(\frac {x^2 + y^2}{2} + xy,$ $ \frac {x^2 + y^2}{2} - xy)$ And so : $ P(x + y,x - y) = P(x + \sqrt {x^2 - y^2},x - \sqrt {x^2 - y^2)}$ $ \forall x,y$ such that $ x\geq |y|$ Let then $ Q(x,y) = P(x + y,x - y)$ We have $ Q(x,y) = Q(x,\sqrt {x^2 - y^2})$ $ \forall x,y$ such that $ x\geq |y|$ So $ Q(x,y) = Q(x, - y)$ $ \forall x,y$ such that $ x\geq |y|$ and so $ \forall x,y$ ($ Q$ is a polynomial). So $ \exists$ $ R(x,y)$ such that $ Q(x,y) = R(x,y^2)$ And : $ R(x,y^2) = R(x,x^2 - y^2)$ $ \forall x,y$ such that $ x\geq |y|$ and so $ \forall x,y$ ($ R$ is a polynomial). And a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial. And so a general solution of initial equation : $ P(x,y) = H(\frac {x + y}{2},(\frac {x - y}{2})^2) + H(\frac {x + y}{2},(\frac {x + y}{2})^2 - (\frac {x - y}{2})^2)$ Which may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$)

we have $ P(x^{2},(-y)^{2})=P(x^{2},y^{2})=P(\frac{(x+y)^{2}}{2},\frac{(x-y)^{2}}{2})=P(\frac{(x-y)^{2}}{2},\frac{(x+y)^{2}}{2}) $. so $ P(x,y) $ is symmetric. then we can write $ P $ in the form $ P(x,y)=Q(x+y)+xyR(x,y) $. then we obtain that $ R(x,y)=\frac{(x-y)^{2}}{4}S(x,y) $ with $ S $ satisfying the initial equation. how can I finish the problem now?

pco wrote: khashi70 wrote: Find All Polynomials $ P(x,y)$ such that for all real $ x,y$ we have : $ P(x^{2},y^{2}) = P(\frac {(x + y)^{2}}{2},\frac {(x - y)^{2}}{2})$ And a general solution of such an equation is $ R(x,y) = H(x,y) + H(x,x^2 - y)$ With $ H(x,y)$ any polynomial. And so a general solution of initial equation : $ P(x,y) = H(\frac {x + y}{2},(\frac {x - y}{2})^2) + H(\frac {x + y}{2},(\frac {x + y}{2})^2 - (\frac {x - y}{2})^2)$ Which may be written in a simplier manner : $ P(x,y) = H(x + y,(x - y)^2) + H(x + y,4xy)$ (for any polynomial $ H(x,y)$) how can we get the general solution ?????

TheBeatlesVN wrote: how can we get the general solution ????? We have to solve $R(x,y)=R(x,x^2-y)$ $\forall x,y$ 1) Any $R(x,y)=H(x,y)+H(x,x^2-y)$ is a solution, whatever is $H(x,y)$ Trivial result : just check 2) any such polynomial may be written as $R(x,y)=H(x,y)+H(x,x^2-y)$ for some $H(x,y)$ Trivial result : just set $H(x,y)=\frac 12R(x,y)$ And so $R(x,y)=H(x,y)+H(x,x^2-y)$ indeed is a general solution for such an equation. Q.E.D.

The answer is $$ P(x,y) \equiv Q \left( x+y, xy(x-y)^2 \right) $$for any $Q(x,y) \in \mathbb R[x,y]$. First we prove all these solutions work. Note that both $$ P(x) \equiv x+y \qquad , \qquad P(x) \equiv xy(x-y)^2 $$satisfy the given assertion. It follows all specified solutions indeed work. Now we prove the converse direction. Changing $y$ to $-y$, it isn't hard to get $f$ is symmetric, i.e. $$ f(x,y) = f(y,x) ~~ \forall ~ x,y \in \mathbb R $$Note if we subtract from $P$ any polynomial $A(x+y)$, then $P$ would still satisfy the assertion. By doing this, we may WLOG assume $P(x,0) \equiv 0$. As $P$ is symmetric, hence $xy$ divides $P(x,y)$. Now putting $y=x$ gives $$ P(x^2,x^2) = P(2x^2,0) = 0 $$It follows $P$ is divisible by $(x-y)$. As $P$ is symmetric, thus $P$ is divisible by $(x-y)^2$. So $xy(x-y)^2$ divides $P$. Again, we cancel a factor of $xy(x-y)^2$ from $P$. Then $P$ still satisfies the given assertion. We keep proceeding like this, and we are done. $\blacksquare$ Remark: I am not exactly sure how this solution set is same as the one mentioned by pco , though I cannot find any flaw in my solution (maybe someone else can comment on this).

guptaamitu1 wrote: I am not exactly sure how this solution set is same as the one mentioned by pco , though I cannot find any flaw in my solution (maybe someone else can comment on this). I did not check your proof/result but it is very common that many totally equivalent general form for a given F.E. exist. As soon as each is a general solution (any function in this form is a solution and any solution may be written in this form), they are quite equivalent. Example with the "trivial" FE : "find all the functions from $\mathbb R\to\mathbb R$ such that $f(x)=f(-x)\quad\forall x$" Different general solution : 1) $g(x)$ whatever is even function $g(x)$ from $\mathbb R\to\mathbb R$ 2) $g(x)+g(-x)$ whatever is function $g(x)$ from $\mathbb R\to\mathbb R$ 3) $g(|x|)$ whatever is function $g(x)$ from $\mathbb R_{\ge 0}\to\mathbb R$ 4) $\forall x\ge 0$ $f(x)=g(x)$ and $\forall x<0$ $f(x)=g(-x)$ whatever is function $g(x)$ from $\mathbb R_{\ge 0}\to\mathbb R$ But for example $g(x)g(-x)$ whatever is $g(x)$ from $\mathbb R\to\mathbb R$ is not a general solution (some solutions can not be written in this form)