Hopefully, this works.

Posted before (for reals though but.........)

If there is some $a$ such that $f(a)=0$ then we have $a,y\implies f(y)=f(y)+a^2\implies a=0$. We have $f(xf(x)+y)=f(y)+x^2\implies f(nxf(x)+y)=f(y)+nx^2$ for integer $n$. Now, for any $a,b\ne 0$ pick integers $n$ and $m$ such that $naf(a)=mbf(b)$ and $mn\ne 0$. Thus, $f(y)+mb^2=f(y+mbf(b))=f(y+naf(a))=f(y)+na^2\implies mb^2=na^2\implies \frac{f(a)}{f(b)}=\frac{a}{b}$. Fixing $b=1$, we get $f(a)=af(1)$. Thus, we have that $x^2f(1)^2+f(1)y=f(1)y+x^2\implies f(1)\in \pm 1$. Also, we get that $f(f(1))=1+f(0)=f(1)^2=1\implies f(0)=0$. Now if $f(1)=1$, we get $f(x)=x$ and if $f(1)=-1$, we get $f(x)=-x$. It's easy to see that both of these solutions work.

Rg230403 wrote: If there is some $a$ such that $f(a)=0$ then we have $a,y\implies f(y)=f(y)+a^2\implies a=0$. And what if no such $a$ exists ?

pco wrote: Rg230403 wrote: If there is some $a$ such that $f(a)=0$ then we have $a,y\implies f(y)=f(y)+a^2\implies a=0$. And what if no such $a$ exists ? I never used the existence of such an $a$, I just said if something goes to $0$ then it is $0$ so that I can pick integer $m,n$ such that $maf(a)=nbf(b)$ for $a,b\ne 0$.

Let $f(1) = c$ and $f(0) = k$. By simple induction, we get $$f(nxf(x)) = nx^2 + k$$for all natural numbers $n$ and rational numbers $x$. Plugging in $x=1$, we get $$f(nc) = n+k$$for all naturals $n$. Substituting $x=nc$ and $y=0$ in the main equation, we get $$f(ncf(nc)) = n^2c^2+k \implies f(n(n+k)c) = n^2c^2+k$$ There are infinitely many naturals $n$ such that $n(n+k)$ is a natural number. For any such $n$, we know from our second equation that $f(n(n+k)c) = n(n+k)+k$. Therefore, $$n(n+k)+k = n^2c^2+k \implies n(c^2-1) = k$$for infinitely many naturals $n$. This forces $k=0$ and $c = \pm 1$. After this, it is straightforward to conclude that the solutions are $f(x)=x$ for all rationals $x$ and $f(x)=-x$ for all rationals $x$.

Rg230403 wrote: I never used the existence of such an $a$, I just said if something goes to $0$ then it is $0$ so that I can pick integer $m,n$ such that $maf(a)=nbf(b)$ for $a,b\ne 0$. You're right. Sorry for my too-quick reading

anser wrote: Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation \[f(xf(x)+y) = f(y) + x^2\]holds for all rational numbers $x$ and $y$. Here, $\mathbb{Q}$ denotes the set of rational numbers. Nice problem Solution- We will prove that $f(x)=x$ and $f(x)=-x$ are the only solutions, note that these indeed satisfy given condition. Claim - - $f$ is surjective. Proof - Note from condition that if $r$ is in range then so is $r+nx^2$ for any integer $n$. Let $s$ be any rational , let $s-f(0)=\frac{p}{q}$ , then $s = f(0) + pq\times (\frac{1}{q})^2$ and hence $s$ lies in range. Now if $f(x)=0$ (cause surjective such $x$ exists) , we get $f(y)=f(y)+x^2 \implies x=0$ and hence $f(0)=0$ $y=0$ gives $f(xf(x))=x^2, x=1 \implies f(f(1))=1$ and $x=f(1) \implies f(f(1))=f(1)^2$ Hence $f(1) \in \{-1,1\}$ If $f(1)=1$ , $x=1 \implies f(y+1)=f(y)+1 \implies f(n) = n$ for all integers $n$ Repeatedly using given condition gives $f(nxf(x))=nx^2$ , now take $n$ such that $nxf(x)$ is integer. So $nx^2=f(nxf(x))=nxf(x)$ , hence $f(x)=x$ for all $x$. Similarly $f(1)=-1$ gives $f(x)=-x$ for all $x$ Hence Proved.

Storage: Let $P(x,y)$ denote the given proposition. Claim: $f(t)=0$ $\implies$ $t=0$. Proof: $P(t,0)$. $\blacksquare$ Applying $P$ repeatedly, we get $f(y+nxf(x))=f(y)+nx^2$ for all positive integers $n$. Replacing $y$ by $y-nxf(x)$ in the above, we get that the expression in fact holds for all integers $n$. Call it $Q(x,y,n)$. Now we claim that $f$ is injective. Indeed, assume that $f(x)=f(y)$ for some rational m=numbers $x,y$. If one of $x,y$ is zero, we are done by the Claim, so assume otherwise. Then we can choose positive integers $m,n \neq 0$ such that $mx=ny$. Then $Q(x,0,m)$ and $Q(y,0,n)$ give $$mx^2 +f(0)=f(mxf(x))=f(nyf(y))=ny^2+f(0)$$which gives $x=y$ as required. Now $Q(x,0,m^2)$ and $Q(mx,0,1)$ give $$f(m^2xf(x))=m^2x^2+f(0)=f(mxf(mx))$$$\implies$ $f(mx)=mf(x)$ for all rational $x$ and integers $m$ by injectivity. In particular, we get $f \left(\frac{1}{n} \right) = \frac{f(1)}{n}$ for all integers $n \neq 0$. $$\implies f \left(\frac{m}{n} \right) =m f \left(\frac{1}{n} \right) = f(1) \cdot \frac{m}{n}$$$\implies$ $f(x)=xf(1)$ for all rationals $x$. Putting this in the original equation, we can easily get $f(1) =\pm 1$, so the only solutions are $$\boxed{f(x)=x \ \ \forall x \in \mathbb{Q}}$$$$\boxed{f(x)=-x \ \ \forall x \in \mathbb{Q}}$$$\blacksquare$

$P(1, 1)$ gives us $$ f(f(1)+1) = 1+f(1) $$which means $a = f(1) + 1$ is a fixed point. We have the following two cases. Case 1: If $a \neq 0$, then $P(a, y)$ gives us $$f(a^2 + y) = a^2 + f(y)$$and for all integers $n$ $$f(a^2n + y) = a^2n + f(y)$$Now, fix $x$ and let $n \neq 0$ be an integer such that both $a^2n^2$ and $n(x + f(x))$ are also integers. Then $P(x + a^2n, y)$ gives us \begin{align*} (x + a^2n)^2 + f(y) &= f((x + a^2n)f(x + a^2n) + y)\\ &= f((x + a^2n)(f(x) + a^2n) + y)\\ &= f(xf(x) + y + a^2(a^2n^2 + n(x + f(x))))\\ & = f(xf(x) + y) + a^2(a^2n^2 + n(x + f(x))) \\ & = x^2 + f(y) + a^4n^2 + a^2n(x + f(x)) \end{align*}and this gives us $f(x) = x$. Case 2: If $a = 0$, then $f(1) = -1$ and $P(1, y)$ gives us $$f(y + 1) = f(y) - 1$$This can be extended to $$f(y + n) = f(y) - n$$for all integers $n$. Again, fix $x$ and let $n \neq 0$ be an integer such that $n(f(x) - x)$ is also an integer. Then \begin{align*} (x + n)^2 + f(y) &= f((x + n)f(x + n) + y)\\ &= f((x + n)(f(x) - n) + y)\\ &= f(xf(x) + y - n^2 + n(f(x) - x))\\ & = f(xf(x) + y) + n^2 - n(f(x) - x) \\ & = x^2 + f(y) + n^2 - nf(x) + nx \end{align*}and this gives us $f(x) = - x$

If we consider the equation for $x=1$, we obtain the equality $f(y+f(1))=f(y)+1$. Inductively, $f(nf(1))=n+f(0)$ for every positive integer $n$. Now let $x=nf(1)$ for a positive integer $n$, and let $y=0$. Then $f(nf(1)f(nf(1)))=f(nf(1)(n+f(0)))=f(0)+n^2f(1)^2$. Now if we pick $n$ such that $n(n+f(0))$ is a positive integer, we obtain the equality $$n(n+f(0))+f(0)=n^2f(1)^2+f(0).$$Since this holds for infinitely many $n$, we have $f(1)^2=1$ and $f(0)=0$. Suppose $f(1)=1$. Inductively, $f(n)=n$ for every nonnegative integer $n$. Also, $f(x+n)=f(x)+n$ for every rational $x$ and every positive integer $n$. From here we can obtain $f(-n)=-n$ for every positive integer $n$. Now, fix $x$, and let $n$ be an integer such that $n(x+f(x))$ is a positive integer. Consider the starting equation for $(x+n, 0)$. On one hand, we have $f((x+n)f(x+n))=(x+n)^2$. On the other hand, $$f((x+n)f(x+n))=f(xf(x)+n(x+f(x))+n^2)=f(xf(x))+n(x+f(x))+n^2=x^2+n(x+f(x))+n^2.$$But then $(x+n)^2=x^2+n(x+f(x))+n^2$, which implies $f(x)=x$. This holds for all $x$, and it's easy to see that the identity function is a solution. On the other hand, the case when $f(1)=-1$ is completely symmetric, and in this case the function $x \mapsto -x$ is a solution.

Set $f(0)=a$. Substitution $y:=-xf(x)$ gives $f(-xf(x))=a-x^2$ for all $x$. (*) Now set $x:=\frac{a-1}{2}, y:=-\frac{a+1}{2}f\left(\frac{a+1}{2}\right)$. Then, using (*), we get $f(xf(x)+y)=f(y)+x^2=a-\left(\frac{a+1}{2}\right)^2+\left(\frac{a-1}{2}\right)^2 = 0$. Hence there exists $c$ with $f(c)=0$. $x:=c, y:=c$ gives $c=0$. Hence $f(x)=0 \iff x=0$. Easy induction yields $f(kxf(x)+y)=y+kx^2$ for $k\in \mathbb Z$. Hence $f(k^2xf(x)+y)=f(y)+k^2x^2=f(y)+(kx)^2=f(kxf(kx)+y)$ for all $k\in\mathbb Z$ and letting $y:=-kxf(kx)$ we obtain $f(k^2xf(x)-kxf(kx))=f(0)=0$, i.e. $k^2xf(x)-kxf(kx)=0$. This simplifies to $kf(x)=f(kx)$ whenever $k\in\mathbb Z$. Then it is routine to obtain $f(x)=xf(1)$ for all $x\in \mathbb Q$ and checking all functions of the form $f(x)=tx$ we find that the only ones that work are $f(x)=x$ and $f(x)=-x$.

Wait, does this work? Taking $y \rightarrow y-xf(x)$ in the given, $f(y)=f(y-xf(x))+x^2$, hence $$f(y-xf(x))+f(y+xf(x))=f(y)-x^2+f(y)+x^2=2f(y),$$therefore $f(a)+f(b)=2f(\frac{a+b}{2})$ for all $a,b \in \mathbb{Q}$, hence $f$ satisfies Jensen's FE. Since we are working over $\mathbb{Q}$, though, $f$ must be linear, hence after substituting to the given equation we easily conclude that $f(x) \equiv x$ or $-x$. Both these functions obviously satisfy, hence we are done. @below Oops. I am dumb.

Orestis_Lignos wrote: Wait, does this work? Taking $y \rightarrow y-xf(x)$ in the given, $f(y)=f(y-xf(x))+x^2$, hence $$f(y-xf(x))+f(y+xf(x))=f(y)-x^2+f(y)+x^2=2f(y),$$therefore $f(a)+f(b)=2f(\frac{a+b}{2})$ for all $a,b \in \mathbb{Q}$, hence $f$ satisfies Jensen's FE. Since we are working over $\mathbb{Q}$, though, $f$ must be linear, hence after substituting to the given equation we easily conclude that $f(x) \equiv x$ or $-x$. Both these functions obviously satisfy, hence we are done. You can’t say this because the diffrence of |a-b|will be 2xf(x) and we don’t know if this is surjective thus we don’t know if we can use all pairs (a,b) where a and b are rational for example let’s assume xf(x) cannot be equal to 5, if so we cannot use the pairs (1,11) or (2,12) since it will mean there exists x such that xf(x)=5.

I think I have a solution that presents new ideas from the ones posted before, but please correct me if I've doubled your idea The beginning is similar to ideas from the solution by square_root_of_3.

anser wrote: Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation \[f(xf(x)+y) = f(y) + x^2\]holds for all rational numbers $x$ and $y$. Here, $\mathbb{Q}$ denotes the set of rational numbers. Let $P(x,y)$ the assertion of the given equation: $P(c,y)$ ($f(c)=0$). $$f(cf(c)+y)=f(y)+c^2 \implies f(y)=f(y)+c^2 \implies c^2=0 \implies c=0$$$P(1,0)$ $$f(f(1)+0)=f(0)+1 \implies f(f(1))=1$$$P(f(1), 0)$ $$f(f(1)f(f(1))+0)=f(0)+f(1)^2 \implies f(f(1))=f(1)^2=1 \implies f(1)=1 \; \text{or} \; f(1)=-1$$If $f(1)=1$: $P(1,x)$ $$f(x+1)=f(x)+1 \implies f(x)=x \; \text{for all} \; x \in \mathbb Z$$Its easy to prove that: $f(axf(x))=ax^2$ for any integer $a$. Take $a=b \cdot c$ and $x=\frac{1}{c}$ $f(bf(\frac{1}{c}))=\frac{b}{c}$ now take $b=1$ to get: $f(f(\frac{1}{c}))=\frac{1}{c}=c \cdot f(\frac{1}{c})$. And that means $f(\frac{1}{c})=\frac{1}{c}$ since we are taking possitive integers... Now we also have that: $$f(\frac{b}{c})=\frac{b}{c} \implies f(x)=x \; \text{for all} \; x \in \mathbb Q$$Actualy work with the same way to get: $f(x)=-x$ for all $x \in \mathbb Q$ if $f(1)=-1$. List of solutions: $f(x)=x$. $f(x)=-x$. For all $x \in \mathbb Q$ thus we are done .

Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that the equation \[f(xf(x)+y) = f(y) + x^2\]holds for all rational numbers $x$ and $y$. Pretty fun . I tried fixed points (don't know why) at first but found the following solution immediately after half way through. The answers are only $f(x) = x$ and $f(x) = -x$, since \[ f(xf(x) + y) = f(x^2 + y) = x^2 + y = x^2 + f(y) \ \text{and} \ f(xf(x) + y) = f(-x^2 + y) = x^2 - f(y) = x^2 + f(y) \]Now, we will prove that there are no other solutions. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation. Claim 01. $f$ is surjective. Proof. First, note that from $P(1,y)$ we get $f(y + f(1)) = f(y) + 1$, and therefore we conclude that if $a \in \text{Im}_f$, then $a - 1 \in \text{Im}_f$. Now, notice that $P(a,bf(b)+cf(c)+df(d))$ gives us \begin{align*} f(af(a) + bf(b) + cf(c) + df(d)) &= a^2 + f(bf(b) + cf(c) + df(d)) \\ &= a^2 + b^2 + f(cf(c) + df(d)) \\ &= a^2 + b^2 + c^2 + f(df(d)) \\ &= a^2 + b^2 + c^2 + d^2 + f(0) \end{align*}We claim that for any rational number $x \ge f(0)$, then there exists $a,b,c,d \in \mathbb{Q}$ such that $a^2 + b^2 + c^2 + d^2 + f(0) = x$. To do this, suppose that $x - f(0) = \frac{k}{\ell} \ge 0$. We know that by Lagrange's Four Square Theorem, there exists $p,q,r,s \in \mathbb{N}_0$ such that $p^2 + q^2 + r^2 + s^2 = k \ell$. Therefore, take $a = \frac{p}{\ell}, b = \frac{q}{\ell}, c = \frac{r}{\ell}, d = \frac{s}{\ell}$, and we are done. Therefore $f$ is surjective at $[f(0), \infty)$. Since we have $a \in \text{Im}_f \Rightarrow a - 1 \in \text{Im}_f$, it follows that $\text{Im}_f = \mathbb{R}$. Claim 02. $f(0) = 0$. Proof. By surjectivity, there exists $a \in \mathbb{Q}$ such that $f(a) = 0$. $P(a,y)$ gives us \[ f(y) + a^2 = f(af(a) + y) = f(y) \Rightarrow a = 0 \] Claim 03. $f(1) = 1$ or $f(-1) = 1$. Proof. Note that $P(x,0)$ gives us $f(xf(x)) = x^2$. Let $a \in \mathbb{Q}$ such that $f(a) = 1$. $P(a,xf(x))$ gives us \[ a^2 + x^2 = a^2 + f(xf(x)) = f(af(a) + xf(x)) = f(a + xf(x)) = x^2 + f(a) = x^2 + 1\]Therefore, $a = -1$ or $a = 1$. Case 01. $f(1) = 1$. $P(1,y)$ gives us $f(y + 1) = f(y) + 1$. It's hence easy to conclude that $f(x+n) = f(x) + n$ for all $n \in \mathbb{Z}$. We'll now prove that $f(x) = x$ for all $x \in \mathbb{Q}$. Since $f: \mathbb{Q} \to \mathbb{Q}$, $f(x) + x$ must be a rational number. Take a value $n \in \mathbb{N}$ such that $n(x + f(x)) \in \mathbb{Z}$, and notice that since $f(xf(x)) = x^2$ for all $x \in \mathbb{Q}$. \begin{align*} f((x+n)f(x+n)) &= (x + n)^2 \\ f((x + n)(f(x) + n))&= (x + n)^2 \\ f(xf(x) + n(x + f(x)) + n^2) &= (x + n)^2 \\ f(xf(x)) + n(x + f(x)) + n^2 &= x^2 + 2nx + n^2 \\ x^2 + n(x + f(x)) + n^2 &= x^2 + 2nx + n^2 \\ nf(x) &= nx \\ f(x) &= x \end{align*} Case 02. $f(-1) = 1$. $P(-1,y)$ gives us $f(y - 1) = f(y) + 1$. It's hence easy to conclude that $f(y - n) = f(y) + n$ for all $n \in \mathbb{Z}$. We'll now prove that $f(x) = -x$ for all $x \in \mathbb{Q}$. Since $f: \mathbb{Q} \to \mathbb{Q}$, $f(x) - x$ must be a rational number. Take a value $n \in \mathbb{N}$ such that $n(x - f(x)) \in \mathbb{Z}$ and notice that since $f(xf(x)) = x^2$ for all $x \in \mathbb{Q}$. \begin{align*} f((x - n)f(x - n)) &= (x - n)^2 \\ f((x - n)(f(x) + n)) &= (x - n)^2 \\ f(xf(x) + n(x - f(x))- n^2) &= (x - n)^2 \\ f(xf(x)) - n(x - f(x)) + n^2 &= x^2 - 2nx + n^2 \\ x^2 - nx + nf(x) + n^2 &= x^2 - 2nx + n^2 \\ nf(x) &= -nx \\ f(x) &= -x \end{align*}and we are done.

Claim 1. There exists some value of $x$ such that $f(x) = 0$ Proof: Assume to the contrary that there is no such value, then since $xf(x) + y$ can achieve any value over the rationals, we are forced that $f$ is never a perfect square, because if it was, we could just choose sufficient $x,y$ such that $f(xf(x) + y) = x^2$, this then forces $f(y) = 0$, contradiction. Then if $f$ is never a perfect square, we also must have $f(y)$ is never a difference of two squares. Now we show that every rational number is a difference of two rational squares. Take $f(y) = \frac pq$, and we desire to solve $\frac pq = (\frac ab)^2 - (\frac cd)^2$, or $16pq = m^2 - n^2$ which always has solutions since we can factor $(m + n)(m -n ) = 8p \cdot 2q$ and then solve a system of equations. Claim 2. $f$ is injective at $0$, and $f(0) = 0$. Proof: Take any value such that $f(x) = 0$, we then have $P(x,y)$ gives $f(y) = f(y) + x^2$, giving $x = 0$. Claim 3. $f$ is surjective. Proof: First notice $y = 0$ gives that $f$ can achieve any perfect square value. We can write $f(xf(x) + y) - x^2 = f(y)$. Since we can have any perfect square as $f(xf(x) + y)$, using the lemma proved in Claim 1 finishes. Claim 4. $f(x) = x$ or $f(x) = -x$. Proof: Notice by induction we have $f(axf(x) )= ax^2$ for integer $a,b$. Then for some $a,b$ we must have $axf(x) = byf(y)$. This then forces $ax^2 = by^2$, or $\frac{x^2}{y^2} = \frac{b}{a} = \frac{xf(x)}{yf(y)}$, which we can then write as $xf(y) = yf(x)$. Now we consider some $x$ such that $f(x) = 1$. We then have $f(x + y) = f(y) + x^2$, then $y = 0$ gives $x^2 = 1$. If $f(-1) = 1$, observe that plugging $P(-1,1)$ gives $0 = f(1) + 1$, so $f(1) = -1$, and likewise assuming $f(1) = 1$ gives $f(-1) = -1$. Using both of these with the above identity finishes. It is trivial to verify that both solutions actually do satisfy the FE.

This should work, but why are other solutions so long? Can someone check this? (x,0)\mapsto f(xf(x))=f(0)+x^2, and by induction with f(nxf(x))=f(0)+nx^2, take (x,nxf(x))\mapsto f((n+1)xf(x))=f(0)+(n+1)x^2 (we can also induct down, so n is any integer), meaning f is surjective. Take x_0:f(x_0)=0\stackrel{(x_0,0)}{\mapsto}x_0=0\implies f(kxf(x))=kx^2=kf(xf(x))\stackrel{f(1):=c,(x=1)}{\implies}f(kc)=kf(c), so f is linear; checking, f(x)=x or -x.

Solutions are $f(x)=\pm x$ which obviously work. By induction we get $f(nxf(x)+y)=f(y)+nx^2$ for integer $n.$ Denote this as $P(x,y,n).$ Take $P(\tfrac{1}{n},0,mn)$ to get surjectivity. Set $f(k)=0,$ then take $P(k,y,1)$ to get $k=0.$ Set $f(h)=1,$ take $P(h,0,1)$ giving $h=\pm 1.$ Take $P(-1,1,1)$ so $f(1),f(-1)$ are not both $1.$ Take $P(\tfrac{1}{n},0,n)$ giving $f(f(\tfrac{1}{n}))=\tfrac{1}{n}.$ Take $P(f(\tfrac{1}{n}),y,1)$ to get $f(\tfrac{1}{n})=\pm \tfrac{1}{n}.$ If $f(1)=1$ take $P(\tfrac{1}{n},0,n^2)$ to get $f(\tfrac{1}{n})\ne -\tfrac{1}{n}$ for integer $n,$ likewise for $f(-1)=1.$ Then take $P(\tfrac{1}{n},0,mn)$ giving $f(\tfrac{m}{n})=\tfrac{m}{n}$ for integers $m,n$ when $f(1)=1$ and similarly $f(-\tfrac{m}{n})=\tfrac{m}{n}$ when $f(-1)=1.$ Thus $f(x)=\pm x,$ done.

Denote the assertion as $P(x, y)$. Claim: We have $f(kxf(x)) = kx^2 + f(0)$ for any integer $k$. Proof: Base case when $k = 1$ is obvious taking $P(x, 0)$. For the inductive step, note that taking $P(x, kxf(x))$ implies \[ P((k+1)xf(x)) = P(kxf(x) + xf(x)) = x^2 + f(kxf(x)) = x^2 + kx^2 + f(0) = (k+1)x^2 + f(0) \]Similarly, we can repeat this process for negatives. taking $P(x, -xf(x))$ shows that $f(-xf(x)) = -x^2 + f(0)$, and the induction down is similar. $\square$ Now, this implies that $f$ is surjective. Taking the $u$ such that $f(u) = 0$, then $P(u, 0)$ implies that $u = 0$. Thus, $f(0) = 0$, and so $f(kxf(x)) = kx^2$. This means that $f(kxf(x) + y) = f(kxf(x)) + f(y)$, so if we could just show that $kxf(x)$ ranges through the rationals, we would almost be done. Then it suffices to show that $f$ is injective, because we already know $f(kxf(x)) = kx^2$. Claim: $f$ is injective. Proof: If $f(a) = f(b)$, then there exist integers $m$ and $n$ such that $maf(a) = nbf(b)$; in particular, $ma = nb$. Then, \[ ma^2 = f(maf(a)) = f(nbf(b)) = nb^2 \]This implies $a = b$. $\square$ Thus, $kxf(x)$ ranges through the rationals, and since $f(kxf(x) + y) = f(kxf(x)) + f(y)$, then $f$ satisfies the Cauchy equation. Thus, $f(x) = cx$ for some constant $c$, and since $f(xf(x)) = x^2$, we find that $f(x) = x$ or $f(x) = -x$, both of which work. $\blacksquare$

The solutions are $f(x) = x$ and $f(x) = -x$, which clearly work. We now prove these are the only solutions. Claim: $f$ is surjective. Proof: Let $y = 0$. Then we get $$f(xf(x)) = f(0) + x^2.$$Letting $y = -xf(x)$, we get $$f(-xf(x)) = f(0) - x^2.$$Using induction we get that $f(0) + nx^2$ is in the range of $f$ for any $n \in \mathbb{Z}$. Letting $n = pq$ and $x = 1/q$ with $p, q \in \mathbb{Z}$, we get $f(0) + p/q$ is in the range of $f$. Therefore $f$ is surjective. Now let $f(a) = 0$. Setting $(a, y)$ we get $$f(y) = f(y) + a^2,$$so $a$ = 0. Therefore, $f(0) = 0$. Let $f(b) = 1$. Setting $(b, 0)$ we get $b^2 = 1.$ WLOG, let $b = 1.$ Then $f(1) = 1$. Setting $(1, x)$, we get $$f(x+1) = f(x) + 1,$$so $f(x) = x$ for all $x \in \mathbb{Z}.$ Claim: For positive integers $k$, $f(kxf(x)) = kx^2$. Proof: Use Induction. For the base case we have $f(xf(x)) = x^2,$ as before. Now assume $k-1$ works. Setting $(x, (k-1)xf(x)),$ we get $$f(xf(x) + (k-1)xf(x)) = f((k-1)x) + x^2.$$Simplifying, we get the desired result. Now pick $k$ such that $kxf(x)$ is an integer. Then $$kxf(x) = kx^2,$$so $f(x) = x$. Similarly, the case where $f(1) = -1$ gives $f(x) = -x$. Therefore, these are the only solutions.

We uploaded our solution https://calimath.org/pdf/EGMO2021-2.pdf on youtube https://youtu.be/t6fHUGQ2sCQ.

Denote the assertion as $A(x,y)$. We proceed with the following 2 claims: Claim 1: $f(c) = 0 \iff c=0$. Our function is surjective, so there must exist such $c$. Substituting $A(c,y)$ finishes. ${\color{blue} \Box}$ Claim 2: $f(nxf(x)+y) = f(y)+nx^2$ for all integers $n$. Follows with straightforward induction after noting \[f(xf(x)+y) = f(y)+x^2, \quad f(-xf(x)+y) = f(y)-x^2. \quad {\color{blue} \Box}\] Thus we can substitute $n=k^2$ and $y=-k^2xf(x)$ to get \[f(k^2xf(x)+y) = f(y)+(kx)^2 = f(kxf(kx)+y) \implies f(nx)=nf(x),\] which implies $f(x)=x \cdot f(1)$ over the rationals. Hence we extract our only solutions $\boxed{f(x)=x, \quad f(x)=-x}$. $\blacksquare$

Let the given assertion be $P.$ Consider an arbitrary $r \in \mathbb{Q}$. Then $r = f(0) + \frac{p}{q}$ for integers $p,q.$ Let $x = \frac{1}{q^2}, y = 0.$ We receive, \begin{align*} f((pq)xf(x)) - f(0) &= \sum_{i = 1}^{pq} f(ixf(x)) - f((i-1)xf(x)) = pqx^2 \\ f((pq)xf(x)) &= f(0) + \frac{p}{q} = r. \end{align*}So $f$ is surjective. Now let there be an $a$ s.t $f(a) = 0.$ $P(a,y)$ yields, $a^2 = 0$ so $a = 0.$ There also exists a $b$ s.t $f(b) = 1.$ Then $P(b, 0)$ yields $f(b) = b^2 = 1$ hence $b = \pm 1.$ Therefore, $P(b, y)$ yields $f(y \pm 1) = f(y) + 1$ and we conclude that for $c \in \mathbb{Z}$, $f(c) = \pm c.$ From $P(x, 0)$, we receive $f(xf(x)) = x^2.$ Let $t$ be the denominator of $xf(x).$ Then, \begin{align*} f(txf(x)) - f(0) &= \sum_{i = 1}^{t} f(ixf(x)) - f((i-1)xf(x)) = tx^2 \\ txf(x) &= \pm tx^2 \\ f(x) &= \pm x. \end{align*}Note that there is no point-wise trap since the sign of the result is fixed based upon $f(1)$. Hence the solutions are $f(x) \equiv \pm x.$

Storage: First, notice that if $f(r)=0$ for some $r \in \mathbb{Q}$ then $a=0$ from $P(a,a)$ Hence for any $a,b\in \mathbb{Q^*}$ then there exists two integers $m$ and $n$ such that $naf(a)=mbf(b)$. Combining this the fact that for any integer $k$ then $f(kxf(x)+y)=kx^2+f(y)$ (simple induction) then $P(a,0)$ and $P(b,0)$ gives us: $na^2=mb^2$ Thus $\frac{f(a)}{a}=\frac{f(b)}{b}$ so $fx)=f(1)x \forall x\in \mathbb{Q^*}$ Replacing $f(x)$ by $f(1)x$ implies $f(1)\in \{1,-1\}$ and $P(x,0)$ implies $f(0)=0$ Hence, the only solutions are $f(x)=x$ and $f(x)=-x$ $$\mathbb{Q.E.D.}$$

Let $P(x, y)$ be the assertion. $P(x, 0)$ gives $f(xf(x)) = f(0) + x^2$. We claim $f(kxf(x)) = f(0) + kx^2$ by induction. It's satisfied for $x = 1$, so say it's true for $\{1, ..., n\}$. We have: $$f((n + 1)xf(x)) = f(nxf(x) + xf(x))$$$P(x, nxf(x))$ gives $f(nxf(x) + xf(x)) = f(nxf(x)) + x^2 = f(0) + nx^2 + n = f(0) + (n + 1)x^2$. So our claim is fulfilled for positives. We now proceed with induction in the same method by using $P(x, -xf(x))$ to get $f(-kxf(x) + xf(x)) = -x^2(k - 1) + f(0)$, so our claim is fulfilled for negatives as well. So we have $f(kxf(x)) = kx^2 + f(0)$, let $kx^2 + f(0) = \frac{a}{b}$ as it ranges through the rationals, taking $k = ab$, $x = \frac{1}{ab}$, we have surjectivity. Let $f(\alpha) = 0$, $P(\alpha, \alpha)$ gives $0 = \alpha^2 \implies \alpha = 0$, so $f(0)$ is unique and equal to $0$. We claim injectivity of $f(x)$, let $f(x) = f(y)$, take positive integers, $m, n$, such that $mx = ny$. We have: $$mx^2 = f(mxf(x)) = f(nyf(y)) = ny^2 \implies (mx)*x = (ny)*y \implies x = y$$which implies injectivity, so we can say that $kxf(x)$ encompasses all the rationals (as the function is surjective as well) $$f(kxf(x) + y) = f(kxf(x)) + f(y) \implies f(x) = cx$$Substituting back, we have $c = 1, -1 \implies f(x) = x, -x$.

I claim that the only solutions are $f(x)\equiv x$ and $f(x)\equiv -x$. These two work; \[f(xf(x)+y)=f(x(x)+y)=x^2+y,\]\[f(y)+x^2=y+x^2,\]and \[f(xf(x)+y)=f(x(-x)+y)=x^2-y,\]\[f(y)+x^2=-y+x^2,\]as desired. (1) First, I claim that for all integers $k$, it holds true that \[f(kxf(x)+y)=f(y)+kx^2.\]I will prove this first for the positive integers using induction. Assume that for all positive integers $1\leq m\leq k-1$, it holds true that \[f(mxf(x)+y)=f(y)+mx^2.\]Then, we have that \[f(kxf(x)+y)=f((k-1)xf(x)+y)+x^2=f(y)+(k-1)x^2+x^2=f(y)+kx^2,\]with the base case holding true at $k=1$ by problem requirements. We can additionally prove this for negative integers by inducting the ``other way'', which gives that; \[f(kxf(x)+y)+x^2=f((k+1)xf(x)+y),\]\[\implies f(kxf(x)+y)=f((k+1)xf(x)+y)-x^2=f(y)+(k+1)x^2-x^2=f(y)-kx^2,\]again with base case of $k+1=1$. This proves our claim. I claim that $f(x)$ is bijective. I will prove this in two parts - surjectivity and injectivity. First, I claim that $f$ is surjective. First, by (1), setting $y=0$, we get that \[f(kxf(x)+0)=f(0)+kx^2,\]for all integers $k$. Since every rational number can be expressed as $kx^2$ for some integer $k$ (not necessarily positive) and rational number $x$, we have that, we get that $f(0)+kx^2$ must span over all rational numbers. Therefore $f$ is surjective. Before proving injectivity, we take advantage of the $0$ case. Let $m$ be a number such that $f(m)=0$. We get that \[f(kmf(m))=f(0)+km^2,\]\[\implies f(0)=f(0)+km^2,\]meaning that $m$ must be $0$. Since $f$ is surjective, there must exist a rational number such that $f(m)=0$. Additionally, we just proved that $0$ is the only such number. This implies that $f(0)=0$ and for all $x\neq 0$, $f(x)\neq 0$. Now I claim that $f$ is injective. FTSOC, assume that $\exists a$, $b\in \mathbb{Q}$ such that $a\neq b$, $a$, $b\neq 0$, and $f(a)=f(b)$. Let $m$ and $n$ be integers such that $ma=nb$ (since $a$, $b$ are rational, they must exist). We then get that \[ma^2=f(maf(a))=f(nbf(b))=nb^2,\]\[\implies a=b\]since $ma=nb$ and $f(a)=f(b)$. This is a contradiction to our requirement that $a\neq b$, therefore $f$ is injective (we can ignore the $0$ case, since we just proved that $f(0)=0$ and no other rational number exists so that $f$ maps it to $0$). Now, since $f$ is both surjective and injective, we have that $f$ must be bijective, proving our claim. Finally, remember we had that \[f(kxf(x))=kx^2,\]from the $y=0$ case of (1). Since $kx^2$ spans over all rational numbers (as we mentioned before) and $f$ is bijective, this means that $kxf(x)$ must also span over all rational numbers! This means that (1) holds for all $k\in \mathbb{Q}$, not just $\mathbb{Z}$. Now, replacing $k$ with $k^2$ and utilizing the generalize to $\mathbb{}Q$ version of (1), we get \[f(k^2xf(x)+y)=f(y)+k^2x^2=f(kxf(kx)+y).\]Setting $y$ to $-kxf(kx)$, we get that \[\implies f(k^2xf(x)-kxf(kx))=f(0),\]and since $f$ is bijective, this implies that \[\implies kx(kf(x)-f(kx))=k^2xf(x)-kxf(kx)=0,\]\[\implies kf(x)=f(kx),\]for all $k$, $x\in \mathbb{Q}$. Letting $f(1)=c$, we get that \[f(x)=xf(1)=cx.\]Plugging this into the original FE, we get that \[f(x(cx)+y)=f(y)+x^2,\]\[\implies c^2x^2+cy=cy+x^2,\]\[\implies c^2=1,\]meaning that $c=1$ or $c=-1$. There is no pointwise trap here, since once $f(1)$ is set, every other value is set. This implies that our only possible solutions are $f(x)\equiv x$ and $f(x)\equiv -x$. We already showed that these worked earlier in the solution, finishing the problem.

Notice that, by induction, $$f(n \cdot xf(x) + y) = f(y) + n \cdot x^{2}$$for any integer $n.$ Now, since this is over the rational numbers, given $a,b,$ we can find integers $m,n$ so that $\frac{m}{af(a)} = \frac{n}{bf(b)}.$ Now, we have that $$n \cdot a^{2} = m \cdot b^{2}$$by the equation before. So, we get that $$\frac{a}{b} = \frac{f(a)}{f(b)},$$and hence $f(a)$ is affine. Plugging in $f(x) = cx,$ we find that the only solutions are $f(x) = x, -x.$

The answer is $f(x) = x$ or $f(x) = -x$. Iterating the FE gives us \[f(nxf(x)+y) = f(y) + nx^2\]for all rationals $x$ and $y$ and all integers $n$ (even negative $n$). From this, the surjectivity of $f$ is obvious. Claim: If $f(x) = 0$, then $x=0$. Proof: Let $r$ be a root. Then, taking $P(r,x)$ gives $f(y) = f(y)+r^2$, so $r=0$. So, $f(0) = 0$. Claim: We have $f$ injective. Proof: Suppose $f(a) = f(b)$. Then, letting $nx^2 = -f(a)$ and plugging into the FE, we have \[0 = f(a)-f(b) = f(a-b) - f(0) = f(a-b),\]so $a=b$. If $n_1x_1^2 = n_2x_2^2$, subtracting $P(x_1, y, n_1)$ and $P(x_2, y, n_2)$ \[n_1x_1f(x_1) = n_2 x_2 f(x_2) \implies \frac{f(x_1)}{x_1} = \frac{f(x_2)}{x_2}.\]But for any pair of rationals $(x_1, x_2)$, there exist integers $n_1$ and $n_2$ for which $n_1x_1^2 = n_2x_2^2$, so it follows that $f(x) = cx$.

By induction, it is easy to see that $f(nxf(x) + x) = nx^2 + f(y)$ for some $n\in \mathbb Z.$ Pick $m, n\in \mathbb Z$ such that at most one of $m, n$ are 0, and $naf(a) = mbf(b).$ Clearly, this exists by the given domain/range, and by casework on if $f(a), f(b)$ are 0. Thus, $f(naf(a) + y) = f(y) + na^2,$ while $f(mbf(b) + y) = f(y) + mb^2.$ Thus, $na^2 = mb^2.$ If only one of $m,n$ are $0,$ wlog we assume $m = 0.$ Then, $a = 0.$ Thus, we can consider when $a, b$ are nonzero. We see that $\frac{f(a)}{a} = \frac{f(b)}{b}.$ Thus, $f \equiv cx$ for some constant $c.$ This gives the two solutions $f\equiv x, f\equiv -x.$