No. $P(x)-x$ has root $3+\sqrt{5}$, and since it has integer coefficients, it also has root $3-\sqrt{5}$. Thus $P(x)=Q(x)(x^2-6x+4)+x$. Then plugging in $x=1+\sqrt{3}$ we find $2+\sqrt{3}=Q(1+\sqrt{3})(2-4\sqrt{3})+1+\sqrt{3}$. Hence $Q(1+\sqrt{3})\cdot(2-4\sqrt{3})=1$. Taking the $\mathbb{Z}[\sqrt{3}]$ norm on both sides gives $N(Q(1+\sqrt{3}))\cdot (-44)=1$. This is impossible since $N(Q(1+\sqrt{3}))$ must be an integer.

Let $Q(x)=P(x)-x$, then $Q(x)\in Z[x], Q(1+\sqrt{3})=1, Q(3+\sqrt{5})=0.$ It mean $Q(x)=0\mod x^2-6x+4\to Q(x)=(x^2-6x+4)R(x)$ and $Q(x)=1\mod (x^2-2x-2)$. $x^2-6x-4=-4x-2\mod x^2-2x-2, R(x)=ax+b\mod x^2-2x-2\to Q(x)=(-4x-2)(ax+b)\mod (x^2-2x-2)=-2(5a+2b)x-2(5a+b)\not =1\mod(x^2-x-2$. Not exist.

International Zhautykov Olympiad 2014, Problem 4 (Second day).

Rust's solution is marred by the fact that midway he switches from (the correct) $x^2-6x+4$ to (the incorrect) $x^2-6x-4$, by a sign inversion. But with the correction made, one can continue in a similar way, and reach the same impossibility conclusion.

tenniskidperson3 wrote: Taking the $\mathbb{Z}[\sqrt{3}]$ norm on both sides gives $N(Q(1+\sqrt{3}))\cdot (-44)=1$. This is impossible since $N(Q(1+\sqrt{3}))$ must be an integer. Can someone explain me what does this mean? thanks.

So the norm of a number $a+b\sqrt{3}$ is defined to be $a^2-3b^2$: what you do to get this number is you multiply by its conjugate number $a-b\sqrt{3}$. You can show that this process is multiplicative: $N(a)N(b)=N(ab)$. So what I did was I took the norm on both sides; on one side we had $Q(1+\sqrt{3})\cdot(2-4\sqrt{3})$ and on the other side we had $1$. The norm of the left side, since it's multiplicative, is $N(Q(1+\sqrt{3})\cdot(2-4\sqrt{3}))=N(Q(1+\sqrt{3}))\cdot N(2-4\sqrt{3})=$ $N(Q(1+\sqrt{3}))\cdot (2^2-3\cdot 4^2)=N(Q(1+\sqrt{3}))\cdot(-44)$. On the right side we get just 1 again. But the thing about integer polynomials in $1+\sqrt{3}$ is that you always end up with $a+b\sqrt{3}$, where $a$ and $b$ must be integers. So whatever we get with $Q(1+\sqrt{3})$, taking the norm of it must be an integer. Unfortunately, we also find that it must be equal to $-\frac{1}{44}$, which is impossible. Did that help explain it a bit?

tenniskidperson3 wrote: Did that help explain it a bit? yes. It's clear now. thank you very much

How very interesting... This problem was posted during the olympiad itself, approximately 1 hour after it started. Note that the author has posted three other problems from the olympiad while it was still going. What a shame...

Being from Kazakhstan you probably know what you're talking about. So you vouch that this timestamp of the post corresponds to the Kazakhstan time zone of the very competition?

Yes, it was held in Almaty, GMT+6, 9.00-13.30 on 14th, 15th of January. According to timestamp, the guy has posted this problem (P4, Day 2) at 9.54 am. Other two problems were also posted during the olympiad.

you are right, @amatysten! @wws??! Solution of Problem4, posted during the olympiad.

We have $3+\sqrt{5}$ is a root of $P(x)-x$ $\Rightarrow 3-\sqrt{5}$ is root, too. Therefore, $P(x)-x=(x^{2}-6x-4)Q(x)$ and $Q(x) \in Z[x]$. Similarly $P(x)-x-1=(x^{2}-2x-2)R(x)$. $P(0)-0=-4Q(0)=-2R(0)+1 \Rightarrow 1$ is even, contradiction.

Here's a different solution: Let $ P(x)=a_0 + a_1x + a_2x^2 + ... + a_nx^n $ for some $n$ Then $ P(1 + \sqrt{3}) = a_0 + a_1(1+\sqrt{3}) + a_2(1+\sqrt{3})^2 + ... + a_n(1+\sqrt{3})^n = 2 + \sqrt{3} $ If we expand we get $ 2 + \sqrt{3}=P(1+\sqrt{3})=\alpha + \beta\sqrt{3} $ and $ 3+\sqrt{5}=P(3 + \sqrt{5})=\gamma + \lambda\sqrt{5} $ where $ \alpha = \sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*3^k) $ and $ \gamma = \sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*5^k*3^{i-2k}) $ Recall the fact that if $ a+b\sqrt{3} = c+d\sqrt{3}$ for integers $ a,b,c,d$ then $a=c$ and $b=d$. Then we have $ \alpha=2 $ and $ \gamma=3 $. Now we substract $\alpha$ from $\gamma$ as follows: $ \gamma-\alpha=1=\sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*5^k*3^{i-2k})-\sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*3^k) $ Now observe that RHS is even because for each $ 0<=i<=n$ we can factor $a_i$ out and then for each $ k $ we'll have $ \dbinom{i}{2k}(5^k*3^{i-2k}-3^k) $ which is obviously even. Thus, $ a_i(\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*5^k*3^{i-2k}-\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*3^k)$ is even, and so is $\sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*5^k*3^{i-2k})-\sum_{i=0}^{n}\ a_i (\sum_{k=0; 2k<=i}\ \dbinom{i}{2k}*3^k) $. Therefore, $ \gamma-\alpha=1$ is even, which is obviously a contradiction. So there are no such polynomials.

By generalized CRT, the problem is equivalent to showing that $(x^2-6x+4)$ and $(x^2-2x-2)$ are not comaximal in $\mathbb Z[x]$. One possible way to show this is by chaining $(x^2-6x+4)+(x^2-2x-2)\subseteq (2,x)\subsetneq \mathbb Z[x]$, which is exactly the solution by gev-gev

Let a = 1 + sqrt(3) and b = 3 + sqrt(5). Let Q(x) = P(x) - x. P(x) exists if and only if Q(x) exists Q(0) = C. a^2 = 2a + 2 a^3 = (a^2) (a) = a(2a + 2) = 2a^2 + 2a = 6a + 4 ... a^n = M*a + N and N is even. Q(a) = 1 it means that C is odd. b^2 = 6b - 4 b^3 = (b^2) (b) = 32b - 24 ... b^n = T*a + L and L is even Q(b) = 0 it means that C is even. C can't be both even and odd so P(x) does not exist.