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Given that for k2n (n=4 and k takes all integer values from 1 to n)

If we want to state the sum of all solutions is evenly divisible by n we write:

n

∑ k2n = 0 (mod n)

k=0

Is there an equally succinct way to state (in one sentence), that each solution taken individually, is evenly divisible by n?

John
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  • Solutions of what? I see one "=" and it asserts that n=4. If n=4, isn't 2n=8? Does k2n refer to a sequence of numbers? – Valtteri May 28 '14 at 22:22
  • A summation equation involving k(squared)n is given in the 3rd 4th and 5th lines. It is an example of how we state the sum of all solutions when integral values of k from 1 to n are substituted into the equation. What I need to know is a way to state that each solution taken individually is evenly divided by n. – John May 28 '14 at 22:39
  • In other words: Is there an equally succinct way to state (in one sentence), that each solution taken individually, is evenly divisible by n?

    12n= 0 (mod n) Ex. (4*1) = 4 = 0 (mod n)

    22n = 0 (mod n) Ex. (4*4) = 16 = 0 (mod n)

    32n = 0 (mod n) Ex. (4*9) = 36 = 0 (mod n)

    42n = 0 (mod n) Ex. (4*16) = 64 = 0 (mod n)

     – John May 28 '14 at 22:50
  • Please see this tutorial on MathJax and any $\LaTeX$ tutorial on the web you like. – Ross Millikan May 28 '14 at 22:54

1 Answers1

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If you mean all the squares are divisible by $n$, you can just say $\forall (k)k^2 \equiv 0 \pmod n$. Is that what you are looking for?

Ross Millikan
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