gouthamphilomath wrote: ... there exists $n$ distinct positive integers such that their sum is a pure $2009-$th power and their product is a pure $2010-$th power. Hi ! By CRT you can easily see that we could replace $2009$ and $2010$ with any coprime numbers .

mahanmath wrote: gouthamphilomath wrote: ... there exists $n$ distinct positive integers such that their sum is a pure $2009-$th power and their product is a pure $2010-$th power. Hi ! By CRT you can easily see that we could replace $2009$ and $2010$ with any coprime numbers . Can you outline your proof by chinese remainder theorem?

Let the coprime integers be p and q. Let a and b positive integers such that ap-bq=1. So we only need to prove that there exists n distinct pure (ap)-th powers whose sum is a pure (bq)-th powers. Let ap=x+1 and bq=x. Consider the integers $s^x, (2s)^x, ..., (ns)^x$ where $s=1^x+2^x+...+3^x$, so the sum is $s^{x+1}$. QED

correct me if im mistaken but didnt u just prove that the product is a 2009th power and the sum is a 2010th power instead of vice versa?

Any pure $ap$-th power is also a pure $p$-th power. So we take $a$ and $b$ such that $2009a-2010b=1$, for example $a=2009,b=2008$.

Because we can, assume that the numbers are all $2010$th powers (its pretty clear we lose no strength by doing so). The first reaction to this is $a \mapsto ca$, where $c$ is a constant. The thing that makes this even nicer is when we plug it into our equation, say $(a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010} = x^{2009}$, where all said variables are integers. At the beginning, take each $a_1$, $a_2$ $\cdots$ $a_n$ to be distinct integers, but otherwise place no restriction. Let $ x = (a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010}$. Now since $(a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010} \mid (x)^{2009}$, just multiply each of the $a_n$ and $x$ by $x^{2008}$, and we're done.

va2010 wrote: Because we can, assume that the numbers are all $2010$th powers (its pretty clear we lose no strength by doing so). The first reaction to this is $a \mapsto ca$, where $c$ is a constant. The thing that makes this even nicer is when we plug it into our equation, say $(a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010} = x^{2009}$, where all said variables are integers. At the beginning, take each $a_1$, $a_2$ $\cdots$ $a_n$ to be distinct integers, but otherwise place no restriction. Let $ x = (a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010}$. Now since $(a_1)^{2010} + (a_2)^{2010} \cdots (a_n)^{2010} \mid (x)^{2009}$, just multiply each of the $a_n$ and $x$ by $x^{2008}$, and we're done. Hello, I'm a bit confused as to how distinct $a_1, a_2, \cdots, a_n$ exist. Sorry, but could you elaborate more on that? Thank you in advance!

Just take $1$ thru $n$

Oops, forgot these were distinct integers. Choose $x = 1^{2010} + 2^{2010} + \dots + n^{2010}$ and consider the numbers $i^{2010} \cdot x^{2008 \cdot 2010}$ for $1 \le i \le n$. Then their sum is $x^{2009^2}$ and their product is the product of $2010$-th powers, hence a $2010$-th power, and so they satisfy the condition.

suli wrote: Oops, forgot these were distinct integers. Choose $x = 1^{2010} + 2^{2010} + \dots + n^{2010}$ and consider the numbers $i^{2010} \cdot x^{2008 \cdot 2010}$ for $1 \le i \le n$. Then their sum is $x^{2009^2}$ and their product is the product of $2010$-th powers, hence a $2010$-th power, and so they satisfy the condition. How did you prove that the sum of numbers $i^{2010} \cdot x^{2008 \cdot 2010}$ is $x^{2009^2}$ ?

AmirAlison wrote: suli wrote: Oops, forgot these were distinct integers. Choose $x = 1^{2010} + 2^{2010} + \dots + n^{2010}$ and consider the numbers $i^{2010} \cdot x^{2008 \cdot 2010}$ for $1 \le i \le n$. Then their sum is $x^{2009^2}$ and their product is the product of $2010$-th powers, hence a $2010$-th power, and so they satisfy the condition. How did you prove that the sum of numbers $i^{2010} \cdot x^{2008 \cdot 2010}$ is $x^{2009^2}$ ? because $\displaystyle\sum i^{2010} x^{2008\cdot 2010}=x^{2008\cdot 2010} \sum i^{2010}=x^{2008\cdot 2010}\cdot x=x^{(2009-1)(2009+1)+1}=x^{2009^2}$

Take $a_i = i^{2010} . c^{2010.2008}$ , with $c = \sum_{i=1}^{n} . b_i^{2008}$ , then $\sum a_i = (c^{2009})^{2009}$ and $ \prod a_i = (n!.c^{2008})^{2010}$

I suppose this construction should work?... Let $a_1,a_2,...,a_n$ be the $n$ numbers that we're dealing with. Now, for some positive integer k, let $a_1=(2^{2009 \cdot 2010}+1)^k$ and $a_2=2^{2009 \cdot 2010}(2^{2009 \cdot 2010}+1)^k,a_3=2^{2009 \cdot 2010}(2^{2009 \cdot 2010}+1)^{k+1} , a_4=2^{2009 \cdot 2010}(2^{2009 \cdot 2010}+1)^{k+2},...,2^{2009 \cdot 2010}(2^{2009 \cdot 2010}+1)^{k+n-2}$. Now, as sum is perfect $2009^{th}$ power, thus $(2^{2009 \cdot 2010}+1)^{k+n-1}$ is a $2009^{th}=>2009 \mid k+n-1$. Similarly $2010 \mid k+(k+(k+1)+...(k+n-2)=nk+\frac{(n-1)(n-2)}{2}$. Hence we're done by CRT...

here's another solution that I've just found using the notation that $1^3 + 2^3+ ......+n^3=(1+2+3+...+n)^2$ let $s=1+2+3...n$ , $p=n!$ we claim that the numbers $s^{2007.2010}p^{2007.2009^2}.i^3$ for each $ n \ge i \ge 1$ then their sum is $s^{2007.2010+2}p^{2007.2009^2}$ but $2007.2010+2 \equiv 0$ $mod$ $2009$ their product is $s^{2007.2010}p^{2007.2009^2+3}$ but $2007.2009^2+3 \equiv 0$ $mod$ $2010$ thus we win

mahanmath wrote: gouthamphilomath wrote: ... there exists $n$ distinct positive integers such that their sum is a pure $2009-$th power and their product is a pure $2010-$th power. Hi ! By CRT you can easily see that we could replace $2009$ and $2010$ with any coprime numbers . Indeed its trivial by CRT, take all of the $a_i$'s to be $2010$th powers,say $x=\sum a_i^{2010}$.By CRT there exists a solution to the equations \begin{align*} 2008k+1 \equiv 0 \pmod{2009} \\ 2008kn \equiv 0 \pmod{2010} \end{align*}Now multiply both sides with $x^{2008k}$

This problem is about how sums and products behave under scaling, and this solution is very "just do it" in spirit, and hopefully I explained the motivation well. The idea is to start with just satisfying the product condition, and use CRT to slowly incorporate the sum condition without "breaking" the product condition. We will generate a valid solution as follows. First, take any arbitrary set of $n$ distinct positive integers $a_1$, $a_2$, $\dots$ $a_n$ such that their product is a perfect 2010th power. Now, consider the prime factorization of their sum. For any prime $p$ dividing the sum, if currently its exponent is $-k\pmod{2009}$, multiply all $a_i$ by $p^r$, where $r$ is a positive integer such that $r\equiv k\pmod{2009}$ and $r\equiv 0\pmod{2010}$ (exists by CRT). This scaling makes the exponent of $p$ in the sum divisible by 2009 while not affecting any other exponents, and $r\equiv 0\pmod{2010}$ makes it so that it also does not affect the fact that the product is a $2010$th power. Repeating for all such primes $p$, we are done.