My favorite algebra in this shortlist, although quite easy for an A6.

Mine! Hope that everybody enjoy this problem!

Easier than A4 and A5 in my opinion. My solution is basically the same as the above one.

All $f$ can be characterized as follows: Let $V=Im (f)$ be a $\mathbb{Z}$-module under addition and $f(v+x)=v+f(x)$ for all $v\in V$. Thus for each element of $\mathbb{R} / V$ we can assign (and must find) a representative to be in the kernel of $f$.

There are worse FE A6s. Solved with KurtGodel5.

Not a bad algebra problem (rare lol)

The answer is $\boxed{\frac{n+1}{n}} $ for any nonzero integer $n$. First we show these work. Claim: $f(nf(0)) = (n +1)f(0)$ for any integer $n$. Proof: If the result is true for $k$, then $P(kf(0),0)$ and $P((k-1)f(0),0)$ give that the result is true for $k+1$ and $k - 1$, respectively. Now, we can use the base case $n=0$, and induct upwards and downwards to get that it holds for all integers $n$. $\square$ This claim implies that $q = \frac{n+1}{n}$ for any nonzero integer $n$ works. Now we show that everything fails. For each $q\ne \frac{n+1}{n}$, we will create a function that has no $z$ satisfying $f(z) = qz$. Clearly the function $f(x) = x + 1$ is a counterexample for $q = 1$, so assume $q\ne 1$ now. Since $q \ne \frac{n+1}{n}$, $\frac{1}{q-1}$ is not an integer. Construct a function $f$ such that $f(x+1) = f(x) + 1$ for all reals $x$, and $f(x)$ is also always an integer. Clearly these functions satisfy the problem condition. Claim: We may construct such a function with the additional condition that $f(x) - qx$ is never a multiple of $q-1$ for any real $x$. Proof: This is equivalent to $\frac{f(x) - qx}{q-1}$ is never an integer. Call $x$ bad if $\frac{f(x) - qx}{q-1}$ is an integer. Notice that if $x$ is bad, then $f(x+1) - q(x+1) = f(x) - qx - (q-1)$ and $f(x-1) - q(x-1) = f(x) - qx + (q-1)$, so $x+1$ and $x-1$ are also bad. Therefore, for each bad number $x$, $\{ x \}$ is also bad. Suppose for all functions satisfying the condition, there existed at least one bad $x$. Since $\frac{1}{q-1}$ is not an integer, we can guarantee that $\frac{0 -qx}{q-1}$ and $\frac{1 - qx}{q-1}$ are not both integers, so we can make a new function $g(x)$ such that for any bad $x$, we have $g(\{x\})\in \{0,1\}$, $\frac{g(\{x\}) - q \{ x\}}{q-1}\not\in \mathbb{Z}$, and $g(x) = g(\{x \}) + \lfloor x \rfloor$. Also, $g(x) = f(x)$ for any $x$ that is not bad. Notice that $g(x)$ still satisfies the conditions, and we note that $\frac{g(x) - qx}{q-1}$ can never be an integer, so we have found a counterexample. $\square$ Now choose $f(x)$ such that $f(x) - qx$ is never a multiple of $q-1$. Suppose we had a solution to $f(z) = qz$ for some $z$. Let $z = n + r$, where $n$ is an integer and $0\le r < 1$. We have $f(z) = f(r) + n$, so $f(r) = qz - n = qr + (q-1)n$, so $f(r) - qr = (q-1)n$ is a multiple of $q-1$, which we know isn't possible with this value of $f$. Therefore this $f$ has no solutions to $f(z) = qz$, which means that any $q\ne \frac{n+1}{n}$ fails.

This was also India tst day 4 p2

f(x+f(y))=f(x)+f^2(y) Therefore,f(x)+f^2(y)=f(x)+f(y) f(x)+f(y)=sqrt(f(x)+f(y) We can see that,it is only possible for 0 and 1. f(z)=qz q=f(z)/z z=f(z)/q Therefore,q=z is possible f(z)=1 is possible 1 is a positive rational number It is clear that,q=1 is the only solution we can get. Also,we can have q=0 Therefore,q=0,1 are the possible solutions.

Banglarmanush wrote: $f(x+f(y))=f(x)+f^2(y)$ Ahh? In condition we know that $f(x+f(y)) =f(x) +f(y) $

Substituting $y=0$ gives $f(x+f(0))=f(x)+f(0)$, so $f(nf(0))=(n+1)f(0)$ for all $n\in\mathbb Z$. Therefore, $nf(0)q\neq f(nf(0))=(n+1)f(0)$ implies $q\neq\frac{n+1}n$ for all $n\in\mathbb Z$ and $n\neq0$. Now, we will show that if $q=\frac ab$ for $\gcd(a,b)=1$ and $|a-b|\neq1$, then there exists $f$ satisfying the conditions. define $f(x)=\frac ab(x+k)$ for all integers $0\leq x<a$ such that $k$ satisfies $b\mid x+k$ and $b-a\mid k-1$, which exists by the Chinese Remainder Theorem since $\gcd(b,b-a)=1$. Then, define $f(ma+n)=f(n)+ma$ for $m,n\in\mathbb Z$ such that $0\leq n<a$. Finally, define $f(x)=a\left\lfloor\frac xa\right\rfloor$ for all $x\not\in\mathbb Z$. Suppose $f(x)=\frac abx$ for some $x$. If $x$ is not an integer, then $\frac abx=a\left\lfloor\frac xa\right\rfloor$, so $x=b\left\lfloor\frac xa\right\rfloor$ implies $x$ is an integer. If $x=ma+n$ is an integer, then $\frac abx=ma+\frac ab(n+k)$ implies $\frac ab(x-bm-n-k)=0$, so $m(a-b)=k$, contradiction since $a-b\mid k-1$ and $a-b\mid k$. Now, since $f(y)$ is a multiple of $a$ for all $y$ and $f(x+a)=f(x)+a$ for all $x$, we get $f(x+f(y))=f(x)+f(y)$. Therefore, there exists a function $f$ satisfying the condition if and only if $q=\frac ab$ for some integers $a$ and $b$ satisfying $\gcd(a,b)=1$ and $|a-b|\neq1$.

fe's are sososohard Denote $P(x,y)$ the assertion. From $P(0,0)$ we know $f(f(0))=2f(0),$ and from $P(f(0), 0)$ we know $f(2f(0))=3f(0).$ This gives arise to the following claim: Claim: For every $n \in \mathbb{Z},$ we have $f(nf(0))=(n+1)f(0).$ Proof. Induct on $n.$ Assume for some $k$ that $f(kf(0))=(k+1)f(0)$ holds. Then $P(kf(0), 0)$ implies \begin{align*}f(kf(0)+f(0))&=f(kf(0))+f(0)\\&=(k+1)f(0)+f(0)\\&=(k+2)f(0),\end{align*}as desired. Now $f(nf(0))=(n+1)f(0)=q \cdot nf(0) \implies q=\tfrac{n+1}{n}.$ Henceforth consider $q \neq \tfrac{n+1}{n}.$ Consider the function $f$ such that $f(0)=1,$ $f(k) \in \mathbb{Z}$ and $f(k)+1=f(k+1)$ for all $k \in \mathbb{R}.$ It is easy to see this function satisfies the given constraints. It suffices to check $f(z)=qz,$ for all $z.$ Observe that this holds for all integer $z,$ as $f(z)=z+1.$ Then $z+1=qz,$ but this is a contradiction to $q$ not being in the aforementioend form. Now consider $z \in (0,1).$ We want to pick $f$ such that $f$ such that $f(k)=qk$ for all $k$ of the form $z+ \mathbb{Z}.$ Now, for every integer $n,$ we want \[f(z+n)=f(z)+n=q(z+n)=qz+qn,\]which is equivalent to $\tfrac{f(z)-qz}{q-1}$ being an integer. This is only possible if $\tfrac{1}{q-1}$ is an integer. But this forces $1=n(q-1) \implies q=\tfrac{n+1}{n},$ contradiction. Thus for all $q \neq \tfrac{n+1}{n},$ it is always possible to choose real $z$ such that $f(z) \neq qz.$

Great problem to demonstrate philosophy! Answer is exactly the $q = 1+\frac{1}{n}$ for $n \in \mathbb{Z}$ and $n \neq 0$. We start by constructing a counterexample. Let's impose the automatic conditions that the range of $f$ is the integers and $f(x+1) = f(x) + 1$ for all $x$. These trivialize the FE. Now for each real $s \in [0, 1)$, it suffices to determine $f(s)$ to determine the entire $f$. In particular, to avoid the $f(z) = qz$ condition, we simply need: $x = \frac{f(s) - qs}{q-1} \not \in \mathbb{Z}$ for all $s$. Since we can choose $f(s) \in \mathbb{Z}$ freely, as long as $\frac{1}{q-1} \not \in \mathbb{Z}$ we are done. This eliminates all $q$ which is not in the desired form. Even without the answer claim, at this point your proof fails for a certain set of $q$. Chances are, there is a deeper reason, not just that your proof skills are lacking. Now, by the el classico nesting method, $f(a_1+f(a_2)+f(a_3)+\cdots+f(a_{k+1})) = f(a_1+f(a_2+f(a_3+\cdots + f(a_{k+1}))) \cdots) = f(a_1)+f(a_2)+\cdots+f(a_{k+1})$. Setting $a_1 = a_2 = \cdots = 0$ gives $q = 1 + \frac{1}{k}$ works for all positive integers $k$ since $f(kf(0)) = (k+1)f(0)$. Similarly, setting $a_1 = - kf(0)$ and $a_2 = a_3 = \cdots = 0$ gives $f(-kf(0)) = (1-k)f(0)$, which gives $q = 1 - \frac{1}{k}$ works for all positive integers $k$, finishing the rest of the solution set for $q$.

A bit different approach...

We uploaded our solution https://calimath.org/pdf/ISL2022-A6.pdf on youtube https://youtu.be/TEeJZLUY3bY.

The answer is everything of the form $1+\tfrac{1}{n}$ whenever $n$ is a nonzero integer. Rewrite the assertion as $f(x+y)=f(x)+y$ whenever $y$ is in the range of $f$. By replacing $x$ with $x-y$ this also implies that $f(x-y)=f(x)-y$ whenever $y$ is in the range of $y$. Claim: We have $f(kf(0))=(k+1)f(0)$ for all $k \in \mathbb{Z}$. Proof: This is obviously true for $k=0$, and $k=1$ follows from plugging in $x=0,y=f(0)$. Now, if the claim is true for $k-1$ and $0$, then $f((k-1)f(0)+f(0))=f((k-1)f(0))+f(0)=(k+1)f(0)$, hence the claim is true for $k$ as well; thus by induction the claim is true for all positive integers $k$. Furthermore, if the claim is true for $k+1$ and $0$, then $f((k+1)f(0)-f(0))=f((k+1)f(0))-f(0)=(k+1)f(0)$, so the claim is true for $k$ as well; thus by induction the claim is also true for all negative integers $k$. $\blacksquare$ This implies that all rationals of the form described work. Now fix some rational $q$ not of this form; we construct a function $f$ without any $z \in \mathbb{R}$ satisfying $f(z)=qz$. Suppose the range of $f$ is a subset of $\mathbb{Z}$ and that $f(x+1)=f(x)+1$ for all $x \in \mathbb{R}$; this function clearly satisfies the assertion. Thus we will only look at such functions. It suffices to assign values to $f(r)$ for each $r \in [0,1)$. Fix a choice of $r$. Then $f(r+k)=f(r)+(r+k)-r$ for all $k \in \mathbb{Z}$, so it suffices to find a choice of $f(r)$ such that $qx=f(r)+x-r \iff x=\tfrac{f(r)-r}{q-1}$ has no integer solution. Obviously $q=1$ fails. Otherwise, let $q=\tfrac{a}{b}$ for coprime integers $a,b$. Suppose that $$\frac{n-r}{q-1}=\frac{b(n-r)}{a-b} \in \mathbb{Z}~\forall n \in \mathbb{Z}.$$This would then imply, by subtracting $n$ and $n+1$, that $\tfrac{b}{a-b} \in \mathbb{Z}$: absurd unless $a-b=1$. Thus we recover our original solution curve only, finishing the problem. $\blacksquare$

Note that $P(xf(0),0)$ gives $f((x+1)f(0))=f(xf(0))+f(0)$ and by induction this implies that $f(nf(0))=(n+1)f(0)=\tfrac{n+1}{n} \cdot nf(0)$ so $q=\frac{n+1}{n}$ works for all integers $n\neq 0$. To prove that these are the only solutions, note that if $q$ is not of that form then $\tfrac{1}{q-1}$ is not an integer. Thus, for each $z\in (0,1)$ we can pick an integer $f(z)$ such that $q-1\neq f(z)-qz$. Then, $f(z)+n=q(z+n)$ has no solutions in $n$. After defining the function on $z\in (0,1)$ we define $f(z)=z+1$ for all integers $z$. Then, $P(x,0)$ gives $f(x+1)=f(x)+1$ so $f(z)+n=f(z+n)$. We have \[f(x)=f(\lfloor x\rfloor + \{x\})=f(\{x\})+\lfloor x\rfloor x\neq qx\]The function is well-defined and it satisfies the conditions, since the range is the set of integers.

the if direction and only if directions are both pretty straightforward, the hardest part is essentially guessing the answer The answer is all values $q$ for which $\frac{1}{q-1}$ is an integer. To prove this works, note that $f(f(y))=f(y)+f(0)$ hence if a function exists where $f(z)\neq qz$ then \[qf(y)\neq f(y)+f(0)\implies f(y)\neq \frac{1}{q-1}\cdot f(0).\]Put another way, $\frac{1}{q-1}\cdot f(0)$ is not in the range of $f$. Note that the range of $f$ is closed under addition and subtraction (by comparing $f(x+f(y))$ and $f(y)$) hence $nf(0)$ is in the range of $f$ for all $n\in \mathbb{Z}$, a contradiction. Now we just need a counterexample for the remaining values. For this, we first impose the restriction that $f$ outputs integers only. Then we impose $f(x+1)=f(x)+1$, which is sufficient. At this point we can actually choose values (details are simple divisibility) for rational inputs in $[0,1)$ and we should be done.

IndoMathXdZ wrote: Mine! Hope that everybody enjoy this problem! I co-authored this problem with @IndoMathXdZ, hope everybody had fun doing this! (and from the reactions, seemed like almost everyone of you did!) Maybe a little bit late to declare co-authorship, but here goes!

It is possible to replace $\mathbb{R}$ with any abelian group with the following small modification. If one wants to keep the original formulation, $\mathbb{R}$ can be replaced by $\mathbb{Q}$-vector spaces. Quote: Let $(G, +)$ be an abelian group. Denote by $\mathcal{F}$ the set of all functions $f : G \to G$ such that $f(x + f(y)) = f(x) + f(y)$ for all $x, y \in G$. Find all pairs $(m, n)$ of integers such that for any $f \in \mathcal{F}$, there exists $z \in G$ such that $m f(z) = nz$. Answer. Any pair $(m, n)$ such that $\gcd(m - n, o(g)) \mid m$ for all $g \in G$, where $o(g)$ is the order of $g$. Solution. First for any $f \in \mathcal{F}$, denote $I_f = f(G)$. The given functional equation gives $f(x) - f(y) = f(x - f(y))$ for any $x, y \in G$, so $I_f$ is closed under subtraction and thus is a subgroup of $G$. (It is possible to characterize $f$ based on $I_f$, as mentioned in #5, but we won't go into that.) Now say that a pair $(m, n)$ is $G$-good if for any $f \in \mathcal{F}$, there exists $z \in G$ such that $m f(z) = nz$. Since $f(0) \in I_f$, this gives $f(k f(0)) = f(0) + k f(0) = (k + 1) f(0)$ for all $k \in \mathbb{Z}$. Thus $(m, n)$ is $G$-good if the following holds: \[ \exists k \in \mathbb{Z}, m(k + 1) f(0) = nk f(0) \iff \exists k \in \mathbb{Z}, o(f(0)) \mid m(k + 1) - nk = (m - n)k + m \iff \gcd(m - n, o(f(0)) \mid m. \]That is, if $\gcd(m - n, o(g)) \mid m$ for all $g \in G$, then $(m, n)$ is $G$-good. It remains to prove the converse. Suppose that $(m, n)$ is $G$-good, and fix some $g \in G$. Let $\{z_j \in G : j \in G/\langle g \rangle\}$ be a set of representatives for the cosets of $\langle g \rangle \subseteq G$. Each element of $G$ has a unique representation as $z_j + h$ for some $j \in G/\langle g \rangle$ and $h \in \langle g \rangle$, and $z_j$ projects down to $j$ modulo $\langle g \rangle$. Choose arbitrary integers $m_j$ for each $j \in G/\langle g \rangle$, and define $f : G \to G$ by \[ f(z_j + h) = m_j g + h \quad \forall j \in G/\langle g \rangle, h \in \langle g \rangle. \]By the property of representatives, this function is well-defined, and it is easy to check that $f(G) = \langle g \rangle$ and then $f \in \mathcal{F}$ for any choices of $m_j$. Now let $(m, n)$ be a $G$-good pair. Then taking the above construction, for each $(m_j)_{j \in G/\langle g \rangle}$, there exists $j \in G/\langle g \rangle$ and an integer $k$ such that \[ m(m_j g + kg) = mf(z_j + kg) = n(z_j + kg) \iff k(m - n)g = nz_j - m_j mg. \]That is, the set $S \subseteq G/\langle g \rangle$ of $j$ such that $nz_j - m_j mg \in \langle (m - n) g \rangle$ is non-empty. If we choose $m_j' = m_j + 1$ for each $j \in S$ and $m_j' = m_j$ for $j \notin S$, we also have $nz_j - m_j' mg \notin \langle (m - n) g \rangle$ for some $j \in G/\langle g \rangle$. By definition, for $j \notin S$, this doesn't hold since $m_j' = m_j$, so this condition holds for some $j \in S$. But then for this $j$, we have \[ nz_j - m_j mg, nz_j - (m_j + 1) mg \in \langle (m - n) g \rangle \implies mg \in \langle (m - n) g \rangle. \]After some algebraic manipulation, this yields $\gcd(m - n, o(g)) \mid m$. We are done.