How Can I Solve It!!

Number Theory Level 5

Determine all sequences $(x_{1},....,x_{2011})$ of positive integers such for every positive integer $n$ there is an integer a with

$x_{1}^{n}+2x_{2}^{n}+....+2011x_{2011}^{n}=a^{n+1}+1$ .

Find the sum of all of the entries of all of the sequences.

Note: Suppose the solutions are $(1,2,3...,2011)$ and $(2,4,6,8,...,4022)$ , then your answer will be $1+2+3+...+2011+2+4+6+....+4022=6069198$

The answer is 4066360651.

1 solution

Gandalf The White
Jan 26, 2015

Assume we have such a sequence.Since all numbers of the sequence are fixed, there exist infinitely many primes bigger than the largest member of the sequence.Setting n=p-1, where p is such a prime, we will get that $a+1$ is congruent to $1+2+...+2011$ modulo an infinite number of primes.Thus, a+1 must be equal to that sum.Now we have to prove that if the following is true for every n,

$x_1^n+2.x_2^n+...=1^n+2.a^n+3.a^n+...$

then (1,a,a,a...,a) is the only solution.(The RHS is just a different way of writing $1+a^{n+1}$ )

Im stuck here and cannot prove this...

Suppose $x_{1},x_{2},..,x_{2011}$ be a valid sequence.

For every $n$ ,we have some $y_{n}$ such that $x_{1}^{n}+2.x_{2}^{n}+...+2011.x_{n}^{n}=y_{n}^{n+1}+1$

See that $x_{1}^{n}+2.x_{2}^{n}+...+2011.x_{n}^{n}<(x_{1}+2.x_{2}+..+2011.x_{2011})^{n+1}$

So , $y_{n}$ is a bounded sequence.There is some $y$ such that $y_{n}=y$ for infinitely many $n$

Now,let m be the maximum of $x_{i}$ and write $x_{1}^{n}+2.x_{2}^{n}+...+2011.x_{n}^{n}=t_{m}m^{n}+t_{m-1}(m-1)^{n}+...+t_{1}$

In fact,for sufficiently large values of $n$

$t_{m}m^{n}+t_{m-1}(m-1)^{n}+...+t_{1}-y.y^{n}-1=0$

Now show that $t_{m}=y=m$ and $t_{1}=1$ and all other $t_{i}$ are 0.

Hope this procedure helps. :)

Souryajit Roy - 6 years, 3 months ago

How Can I Solve It!!

The answer is 4066360651.

1 solution

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