# Determine positive integers

If $n\geq3$ is a positive integer, determine $n+1$ positive integers With the property that the sum of all $n$ integers from the $n+1$ numbers build the set: $$S=\{n^2+2,n^2+4,n^2+6,...,n^2+2n-2,n^2+2n\}.$$ For example, if $n=3$, the numbers are $3,3,5,7$.
If $n=4$ the numbers are $2,4,6,6,8$. So for $n=3$ the set $S$ is $11,13,15$. We have $3+3+5=11, 3+5+7=15, 3+7+3=13$. We have four numbers and the sum of $3$ arbitrary picked numbers which I have detmined must be an element of $S$.

The question is: how to detrmine $n+1$ numbers with that property, also $n$ numbers picked from the sum (arbitrary) must be an element of $S$ and all the sums of $n$ numbers must build the set $S$.

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How do you get $3,3,5,7$ for $n=3$. You have to word your problem better. – user44197 Jan 10 '14 at 5:41
What is the question? – user127.0.0.1 Jan 10 '14 at 5:42

Observing the cases for smaller $n$s will give us the following numbers :

When $n$ is odd, the numbers you want are $3,3,5,7,\cdots, 2n-1,2n+1.$

When $n$ is even, the numbers you want are $2,4,6,\cdots, n, n+2, n+2, n+4,\cdots, 2n-2, 2n$.

In the following, let us prove that these are true.

In the case that $n$ is odd, you can easily see $$3+3+5+\cdots+(2n-1)=3+\sum_{k=1}^{n-1}(2k+1)=3+2\cdot\frac{n(n-1)}{2}+n-1=n^2+2,$$ $$3+5+7+\cdots+(2n+1)=\sum_{k=1}^{n}(2k+1)=2\cdot\frac{n(n+1)}{2}+n=n^2+2n.$$

Also, since the difference between any two of consective elements is either $0$ or $2$, you'll see for each $i$ there is a set of $n$ elements whose sum is $$n^2+2i\ \ (i=1,2,\cdots, n).$$

For example, changing $2n-1$ to $2n+1$ in the first equation above will give you $$3+3+5+7+\cdots+(2n-7)+(2n-5)+(2n-3)+(2n+1)=n^2+2+2=n^2+4.$$ Also, changing $2n-3$ to $2n-1$ in the equation above will give you $$3+3+5+7+\cdots+(2n-7)+(2n-5)+(2n-1)+(2n+1)=n^2+4+2=n^2+6.$$ Also, changing $2n-5$ to $2n-3$ in the equation above will give you $$3+3+5+7+\cdots+(2n-7)+(2n-3)+(2n-1)+(2n+1)=n^2+6+2=n^2+8.$$

You can repeat these operations, then finally you'll get $$3+5+7+\cdots+(2n+1)=n^2+2n.$$ Thus, we now know that the numbers above are what you want.

You'll have the numbers for the case that $n$ is even by the same argument above.

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