For which $ n\in \mathbb{N}$ do there exist rational numbers $ a,b$ which are not integers such that both $ a + b$ and $ a^n + b^n$ are integers?
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Tags: function, algebra unsolved, algebra
16.04.2008 07:36
Here is my solution .We only need consider $ n>1$ The first we consider : $ a=\frac{x}{p},y=\frac{y}{q}$ where $ \gcd(x,p)=\gcd(y,q)=1$ and $ p,q>0$ The condition equivalent i)$ pq|xq+py$ From 1 we have $ q|p,p|q$ so $ p=q$ So we only need to consider when $ \gcd(x,y)=1$ Then condition equivalent : i)$ p|x+y$ $ p^n|x^n+y^n$ Case 1 $ n\equiv 0 (\mod 2).$ Because $ \gcd(x+y,x^n+y^n)\in\{1,2\}$ do so $ p=2$ But $ 2^n|x^n+y^n$ then $ 2|x,2|y$ illogic. Case 2 $ n\equiv 1 (\mod 2)$ Call p is a prime divisor of n . Then $ v_p (x^n+y^n)=v_{p}(x+y)+v_{p}(n)$ If we chose $ v_p(x+y)\geq v_{n}(n-1)$ then it satisfies conditon . So answer is $ n\equiv 1 (\mod 2)$
16.04.2008 09:47
What is your function $ v$? Is it the power of $ p$ that divides $ n$? Why $ v_p (x^n+y^n)=v_{p}(x+y)+v_{p}(n)$?
16.04.2008 09:58
To eay you only chose $ p^n|x+y$ This lemma posted many time on Mathlinks
16.04.2008 15:17
TTsphn wrote: This lemma posted many time on Mathlinks Which lemma are you referring to?
16.04.2008 22:05
Hi. I was just wondering where you got that problem from? It was in Croatia TST a week ago, but I'm sure it's from some older competition. I'd just like to know which. I've solved it (Wee!) so I'd like to know where it's from (I actually develop a sort of an emotional bond with the not-so-easy problems I solve on important competitions ). Thanks a lot!
17.04.2008 06:42
Well, mornik, I took it from Croatia TST It's in the imo.org.yu website. Anyway, for the $ n$ odd case, I guess we can just choose $ a=-b$ not being integers. But what if we want positive $ a,b$? Can you help me?
17.04.2008 08:38
Yes chose $ p^n|x+y$ and $ \gcd(x,y)=1$ There exist infinite such pair $ (x,y)$
17.04.2008 09:06
Simply choose $ a = \frac {2^n + 1}{2}$ and $ b = \frac {2^n - 1}{2}$ and you got yourself a solution for odd $ n$.