I don't know whether my solution is correct, as a Problem 4 should not be so easy. It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p-2})$ are of the same parity and they are not equal to zero. So for $ p \ge 3$ being odd, $ (p+1)$ is even $ \Rightarrow 2^{2p-2}|(p+1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x)=x^2-4x+3$,$ x=2$ that satisfies the conditions.

Come back after long inactive time! My answer is p has the form 4*k+1. You could prove that for any polynomial Q and an arbitrary integer number x, we have the product Q(x).Q(x^2)... is congruent module p with a perfect square ( thus groups the 2p-2 terms in pairs). From here we got p divides C^2+1 for certain C. Hence p=4k+1 Now for p has the form 4k+1. We could use the Wilson theorem to give an example

stephencheng wrote: I don't know whether my solution is correct, as a Problem 4 should not be so easy. It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p - 2})$ are of the same parity and they are not equal to zero. So for $ p \ge 3$ being odd, $ (p + 1)$ is even $ \Rightarrow 2^{2p - 2}|(p + 1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x) = x^2 - 4x + 3$,$ x = 2$ that satisfies the conditions. Sorry I have made a typo, it should be $ Q(x),Q(x^2),...,Q(x^{2p - 2})$ are of the same parity . The words ''none of'' should be deleted. Also, mto, it seems that you and I got different results . Can you give an example for $ p=5$?

stephencheng, your result is true. Besides, the fact that this is very easy for a TST problem is true but as I learned from our teachers this solution hadn't recognised before the contest. If we replace p^p by p in the initial question, it may be more interesting or at least closer to the expected solution

I cant understand the problem :s is for everything polinomial ? because if Q(x ) =x, then $ 1+p+x^{(p-1)(2p-1)} = P (x)$, then , there exist $ a$ integer such that P(a)=0, then if p : odd p-1 = even, -> P(x) >0, if p=2, 3+x^3=0, there isnt solution in integer.

aev5peru wrote: I cant understand the problem :s is for everything polinomial ? because if Q(x ) =x, then $ 1 + p + x^{(p - 1)(2p - 1)} = P (x)$, then , there exist $ a$ integer such that P(a)=0, then if p : odd p-1 = even, -> P(x) >0, if p=2, 3+x^3=0, there isnt solution in integer. The problem is : for which primes $ p$ does there exist $ Q(x)$ so that $ 1+p+Q(x^1) \cdot Q(x^2) \cdot ... \cdot Q(x^{2p-2})=0$ has integer root(s)?

it is very easy problem I think it can be good for Junior Balkam M.O.

Why do we have that $ 2^{p-2}|{p + 1}$ ?

Hello Do you mean 'Why do we have that $ 2^{2p - 2}|{p + 1}$ ? It is because $ Q(x),Q(x^2),...$ all have to be even. Hope that helps

stephencheng wrote: I don't know whether my solution is correct, as a Problem 4 should not be so easy. It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p-2})$ are of the same parity and they are not equal to zero. So for $ p \ge 3$ being odd, $ (p+1)$ is even $ \Rightarrow 2^{2p-2}|(p+1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x)=x^2-4x+3$,$ x=2$ that satisfies the conditions. for $p=2$ there exist a polynomial of degree $1$.$Q(x)=2x-5$ and the equation have $x=2$ as a solution.

The only answer is $p=2$. The key lemma is if $x \equiv y \pmod{m}$ then $Q(x) \equiv Q(y) \pmod{m}$. Firstly assume that $p \equiv 3 \pmod{4}$. FTSOC let there be an integer root $x$. Then we have $\prod_{i=1}^{2p-2} Q(x^i) \equiv \prod_{i=1}^{p-1} Q(x^i)^2 \Rightarrow \prod_{i=1}^{p-1} Q(x^i)^2+p+1 \equiv y^2+1 \equiv 0 \pmod{p}$ for some integer $y$. Contradiction since $p \equiv 3 \pmod{4}$. Now assume that $p \equiv 1 \pmod{4}$. FTSOC let there be an integer root $x$. We will reach contradiction by $\pmod{4}$. 1) $x$ is odd. Then we would have $x^k \equiv x^{k+2} \pmod{8} \Rightarrow \prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(x)^{p-1}Q(x^2)^{p-1}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ for some integer $y$. Contradiction since $2$ is not a QR for $\pmod{4}$. 2) $x$ is divisible by $4$ Using $Q(x^i) \equiv Q(0) \pmod{4}$, we have $\prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(0)^{2p-2}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ which is again a contradiction because of the same argument. 3) $x \equiv 2 \pmod{4}$ We have $\prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(0)^{p-1}Q(2)^{p-1}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ which is again a contradiction because of the same argument. For $p=2$ we need $Q(x)Q(x^2)=-3$ for some polynomial $Q$. Trying small cases gives us $x=-1$ is a solution where $Q(x)=2x-1$.

stephencheng wrote: I don't know whether my solution is correct, as a Problem 4 should not be so easy. It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p-2})$ are of the same parity and they are not equal to zero. So for $ p \ge 3$ being odd, $ (p+1)$ is even $ \Rightarrow 2^{2p-2}|(p+1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x)=x^2-4x+3$,$ x=2$ that satisfies the conditions. I don't know how you get from $ (p+1)$ is even to $ 2^{2p-2}|(p+1)$

Because 1+p+Q(x)….=0, so 1+p=-Q(x)….. Here the RHS is divisble by 2^2p-2, so the LHS has to be too