Let $P(x,y)$ denote the statement $f(x+f(y)^2) = f(x+y)^2$ and let $\beta = f(0)^2$. From $P(x,0)$ we get $f(x+\beta) = f(x)^2$, so $f(x) \ge 0$ for all real $x$. Plugging in the original equation we get $f(x+f(y+\beta)) = f(x+y+\beta)$, i.e. $f(x+y) = f(x+f(y))$ for all $x,y \in \mathbb{R}.$ In particular, $f(x+f(0)) = f(x)$ and $f(f(x)) = f(x)$ for all $x$. By induction, $f(x+n \beta) = f(x)^{2^{n}}$ and $f(x+nf(0)) = f(x)$ for all positive integers $n$, and so if $f(0) = p/q$, we have $f(x)=f(x+np^2) = f(x)^{2^{nq^2}}$, i.e. $f(x) \in \{0,1\}$ for all real $x$. If $f$ is identically $0$ or $1$, we have two solutions, otherwise $0, 1$ are both in the image of $f$ and so $f(0)= 0, f(1) = 1.$ So $1 = f(-1 + 2) = f(-1+f(2))$, i.e. $f(2) = 0$ and $1 = f(1/2 + f(1/2))$, i.e. $f(1/2) = 1$ and $f(3/2) = 1$. But then $0 = f(1/2 + 3/2) = f(1/2 + 1)$, contradiction. So the only solutions are $f(x) \equiv 0$ and $f(x) \equiv 1.$ This problem seems very interesting if we drop the condition $f(0) \in \mathbb{Q}$. Are there any other functions such that $f(x+f(0)^2) = f(x)^2$ and $f(x+y)=f(x+f(y))$ for all $x, y \in \mathbb{R}$?

proglote wrote: This problem seems very interesting if we drop the condition $f(0) \in \mathbb{Q}$. Are there any other functions such that $f(x+f(0)^2) = f(x)^2$ and $f(x+y)=f(x+f(y))$ for all $x, y \in \mathbb{R}$? We can slightly generalize the condition to "If there exists $x$ such that $f(x)\in\mathbb{Q}$"

Posted before , check before posting please : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=553244

proglote wrote: Let $P(x,y)$ denote the statement $f(x+f(y)^2) = f(x+y)^2$ and let $\beta = f(0)^2$. From $P(x,0)$ we get $f(x+\beta) = f(x)^2$, so $f(x) \ge 0$ for all real $x$. Plugging in the original equation we get $f(x+f(y+\beta)) = f(x+y+\beta)$, i.e. $f(x+y) = f(x+f(y))$ for all $x,y \in \mathbb{R}.$ In particular, $f(x+f(0)) = f(x)$ and $f(f(x)) = f(x)$ for all $x$. By induction, $f(x+n \beta) = f(x)^{2^{n}}$ and $f(x+nf(0)) = f(x)$ for all positive integers $n$, and so if $f(0) = p/q$, we have $f(x)=f(x+np^2) = f(x)^{2^{nq^2}}$, i.e. $f(x) \in \{0,1\}$ for all real $x$. If $f$ is identically $0$ or $1$, we have two solutions, otherwise $0, 1$ are both in the image of $f$ and so $f(0)= 0, f(1) = 1.$ So $1 = f(-1 + 2) = f(-1+f(2))$, i.e. $f(2) = 0$ and $1 = f(1/2 + f(1/2))$, i.e. $f(1/2) = 1$ and $f(3/2) = 1$. But then $0 = f(1/2 + 3/2) = f(1/2 + 1)$, contradiction. So the only solutions are $f(x) \equiv 0$ and $f(x) \equiv 1.$ This problem seems very interesting if we drop the condition $f(0) \in \mathbb{Q}$. Are there any other functions such that $f(x+f(0)^2) = f(x)^2$ and $f(x+y)=f(x+f(y))$ for all $x, y \in \mathbb{R}$? Nice!

Let $P(x,y)$ stand for $P$lugging in $(x,y)$ in the equation $f(x+f(y)^2 ) = {f(x+y)}^2$. Call $f(0)=q$. Note $P(0,0)$ gives $f(q^2)=q^2$, and then $P(0,q^2)$ gives $f(q^4)=q^4$, and then $P(0,q^4)$ gives $f(q^8)=q^8$, and so on. In general, we have $f(q^{2^k})=q^{2^k}$ for all $k\in\mathbb N$. Now $P(x,q^2)$ says $f(x+q^4)=f(x+q^2)^2=\left(f(x)^2\right)^2=f(x)^4$. Then $P(x,q^4)$ gives $f(x+q^8)=f(x+q^4)^2=f(x)^8$. Iterating this process, we arrive at $f(x+q^{2^k})=f(x)^{2^k}$ for all $k\in\mathbb N$; call this equation $(\star)$. Setting $x\mapsto x+q^{2^k}$ in $(\star)$ gives $$f\left(x+2q^{2^k}\right)=f\left(x+q^{2^k}\right)^{2^k}=\left(f(x)^{2^k}\right)^{2^k}=f(x)^{2^{2k}}.$$Again, performing the replacement $x\mapsto x+q^{2^k}$ in the above equation yields $f\left(x+3q^{2^k}\right)=f(x)^{2^{3k}}$. By induction, it's easy to see that $f\left(x+nq^{2^k}\right)=f(x)^{2^{nk}}$ for all $k\in\mathbb N,n\in\mathbb N_0$. Setting $x=0$ here, we arrive at $f\left(nq^{2^k}\right)=q^{2^{nk}}$ for all $k\in\mathbb N,n\in\mathbb N_0$. Call this result $(\spadesuit)$. Now $P(-f(y)^2,y)$ gives $q=f(\text{something})^2$, so $q\ge 0$. (In fact, $P(x-f(y)^2,y)$ gives $f(x)\ge 0$ for all $x$.) Let's say $q=\tfrac rs$, for $r\in\mathbb N_0, s\in\mathbb N$. Consider the numbers $n_1=r^2s^2,n_2=s^4,k_1=1,k_2=2$. It's easy to check that $n_1q^{2^{k_1}}=n_2q^{2^{k_2}}$. So $(\spadesuit)$ gives $q^{2^{n_1k_1}}=q^{2^{n_2k_2}}$. If $q\not\in \{0,1\}$, then this implies $n_1k_1=n_2k_2\iff r^2s^2\cdot 1=s^4\cdot 2\implies \tfrac rs=\sqrt2$, which is ab-surd (That pun was totally not intended. Not at all.). So in fact $q\in\{0,1\}$. Putting this in $(\spadesuit)$ with $n=1$ gives $f(x+q)=f(x)^{2^k}$, and using this for $k=1$ and $k=2$, we have $f(x)^2=f(x)^4\implies f(x)^2=f(x)$ because $f(x)$ is non-negative. Using this in the original equation, $f(x+f(y))=f(x+y)$. We will call this $Q(x,y)$ (stands for $Q$uoting the given equation with $x,y$). Further, $f(x)\in\{0,1\}$ for all $x$. Now consider two cases: $f(0)=1$. If there is some $x$ with $f(x)=0$, then $Q(0,x)$ gives $f(0)=f(x)=0$, which is bad. So $\boxed{f\equiv 1}$, which is a solution. $f(0)=0$. If there is some $x$ so that $f(x)=1$, then as in Case 1, $f(1)=1$. If $f(\tfrac12)=0$, then $Q(\tfrac12,\tfrac12)$ gives a contradiction, so $f(\tfrac12)=1$, and $Q(\tfrac12,\tfrac12)$ gives $f(\tfrac32)=1$. Then $Q(1,\tfrac12)$ says $f(2)=1$. But according to $Q(1,-1)$, $f(1+f(-1))=0$, and both $f(-1)=1$ and $f(-1)=0$ leads to a contradiction. So in fact $\boxed{f\equiv 0}$, which is another solution. $\blacksquare$

Nice! goldeneagle wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(0) \in \mathbb Q$ and \[f(x+f(y)^2 ) = {f(x+y)}^2.\](25 points) Answer. $f(x) \equiv 0$ and $f(x) \equiv 1$ are the only valid functions. It is easy to check they work. Now we prove that any $f$ validating the assertion $$f(x+f(y)^2)=f(x+y)^2$$for all $x,y \in \mathbb{R}$ with $f(0) \in \mathbb{Q}$ must be either of these two functions. Note that $(x,y) \mapsto (x+f(y)^2)$ is surjective over $\mathbb{R}$, hence $f(x) \ge 0$ for all $x \in \mathbb{R}$. Substitute $y=0$ so $f(x+f(0)^2)=f(x)^2$. Apply $x \mapsto x+f(0)^2$ repeatedly, so $$f(x+Nf(0)^2)=f(x)^{2^N}$$for all $x \in \mathbb{R}$ and any integer $N>0$. Put $x=0$ to obtain $f(f(t)^2)=f(t)^2$ for all $t \in \mathbb{R}$. Then $$f(f(x+f(y)^2))=f(f(x+y)^2)=f(x+y)^2=f(x+f(y)^2)$$for all $x, y \in \mathbb{R}$; hence $f(f(z))=f(z)$ for all $z \in \mathbb{R}$. Substitute $y \mapsto f(z)$ in the original equation; we see $$f(x+z)^2=f(x+f(z)^2)=f(x+f(f(z))^2)=f(x+f(z))^2$$for all $x,z \in \mathbb{R}$. By positivity of elements in $\text{Im} \, f$ we see that $f(x+f(0))=f(x)$ for all $x \in \mathbb{R}$. Now we have $$f(x+Mf(0))=f(x)$$for all integers $M>0$. Pick $M, N$ accordingly, so that $f(0)=\tfrac{M}{N}$ yielding $$f(x)^{2^N}=f(x+Nf(0)^2)=f(x+Mf(0))=f(x)$$hence $f(x) \in \{0,1\}$ for all $x \in \mathbb{R}$. It only remains to eliminate the point-wise trap. Notice that $f(x+f(z))=f(x+z)$ for all $x,z \in \mathbb{R}$ so $$f(f(x)+f(z))=f(f(x)+z)=f(x+z)$$for all $x,z \in \mathbb{R}$. Hence we only need to show $f(0)=f(1)=f(2)$. Suppose $f(0)=1$. Then $f(f(0))=f(0)$ so $f(1)=1$. Also $f(2)=f(1+f(0))=f(1+0)=1$. Hence $f \equiv 1$. Suppose $f(0)=0$. Then if we can find $a \in \mathbb{R}$ with $f(a)=1$; consider $f(f(a))=f(a)$ so $f(1)=1$. Plug $y=1-x$ so $f(x+f(1-x))=f(1)=1$ for all $x \in \mathbb{R}$. If $f\left(\tfrac{1}{2}\right)=0$ then put $x=\tfrac{1}{2}$ to get $f(0)=1$, contradiction! Hence $f\left(\tfrac{1}{2}\right)=1$ so $f(\tfrac{3}{2})=f(1+f(\tfrac{1}{2}))=f(2)$. Now $f(f(2)-1)=1$ so $f(2) \ne 1$ hence $f(2)=0$. But $f(\tfrac{1}{2}+f(\tfrac{3}{2}))=f(2)$ hence we get a contradiction! So $f \equiv 0$. So we have proved that no other functions work, too!

Letting $y=0$ and then $x=0$ gives $$f(x+f(0)^2)=f(x)^2\text{ and }f(f(y)^2)=f(y)^2.$$Applying the first equation repeatedly, we have for all natural $n$ that $f(nf(0)^2)=f(0)^{2^n}.$ Replacing $y$ with $f(y)^2$ and using the second identity gives $$f(x+f(y)^4)=f(x+f(y)^2)^2=f(x+y)^4.$$Iterating this for $y=0$ gives $f(nf(0)^4)=f(0)^{4^n}.$ Suppose $f(0)=p/q$. Then, $$f(p^4)=f(q^4f(0)^4)=f(0)^{4^{(q^4)}}$$But also, $$f(p^4)=f(p^2q^2f(0)^2)=f(0)^{2^{(p^2q^2)}}.$$Hence, it follows that $f(0)\in\{0,1,-1\}$. Actually -1 doesn't work. Indeed, taking $x=-f(y)^2$ in the original FE gives $$-1=f(0)=f(y-f(y)^2)^2,$$which is absurd. So either $f(0)=0$, or $f(0)=1$. If $f(0)=0$, then letting $y=0$, we get $f(x)^2=f(x)$. Note that his means for each $x$, $f(x)\in\{0,1\}$. Hence, the original FE is now $f(x+f(y))=f(x+y)$. We claim there is no $a$ such that $f(a)=1$. Indeed, assume there is. Then, $f(1)=f(f(a)^2)=f(a)^2=1$. Now, $f(0.5+f(0.5))=f(1)=1$. If $f(0.5)=0$, then, we have $f(0.5)=f(1)=1$, not true. So $f(0.5)=1$. But now, consider $f(-0.5+f(0.5))=f(0)=0$, so throwing this in implies $f(0.5)=0$, which is again a contraction. Hence, $\boxed{f(x)=0}$ for all x. Otherwise $f(0)=1$. Putting $y=0$ gives $f(x+1)=f(x+1)^2$, from which we infer for each $x$, $f(x)\in\{0,1\}$. We claim that $f(x)=1$ for all $x$. Indeed suppose there is an $a$ such that $f(a)=0$. Then $f(0)=f(f(a)^2)=f(a)^2=0$, contradcition since we assumed $f(0)=1$. Hence, the other solution is $\boxed{f(x)=1}$, and we're done.