Re: help with number problem

On May 21, 2006, 12:38 AM CT, William Elliot wrote:

On Sun, 21 May 2006, Narcoleptic Insomniac wrote:
On May 20, 2006 10:11 PM CT, isaac_2004@xxxxxxxxx wrote:

hello i am asked to prove that every integer n
for the equation

((n^5)/5) + ((n^3)/3) + (7n)/15

is an integer

You have (3n^5 + 5n^3 + 7n) / 15 and so it must be
shown that

3n^5 + 5n^3 + 7n == 0 (mod 15)

..or equivalently...

3n^5 + 5n^3 + 7n == 0 (mod 3)
..and...
3n^5 + 5n^3 + 7n == 0 (mod 5).

Using a^p = a (mod p), if p prime

Working modulo three we obtain

3n^5 + 5n^3 + 7n == 2n^3 - 2n == 2n^3 - 2n (mod 3).

== 5n + 7n = 12n = 0 (mod 3)

Now 2n^3 - 2n = 2n(n^2 - 1) = 2n(n - 1)(n + 1),
which is clearly divisible by 3 since it is a product
of three consecutive integers. Hence, we have
established

3n^5 + 5n^3 + 7n == 0 (mod 3).

Oh horrors. Maybe he was to prove by induction.

Working modulo five we see that

3n^5 + 5n^3 + 7n == 3n^5 - 3n (mod 5).

== 3n + 7n = 10n = 0 (mod 5)

Note that

3n^5 - 3n = 3n(n^4 - 1) = 3n(n^2 - 1)(n^2 + 1) =
3n(n - 1)(n + 1)(n^2 + 1)

..and since...

Oh horrible horror. Nay, induction is easier than
that.
f(n) = 3n^5 + 5n^3 + 7n
f(-n) = -f(n)
f(0) = 0; f(1) = 15
f(n+1) = f(n) + 15n(...) + f(1) = 15 * ???

Most likely he was to proceed via induction. Granted,
I should have thought to use Fermat's theorem and my
method was a bit long, but I don't see the "horror" in it.

n^2 + 1 == n^2 - 4 (mod 5)

..it follows that...

3n(n - 1)(n + 1)(n^2 + 1) == 3n(n - 1)(n + 1)(n^2 - 4) =
3n(n - 1)(n + 1)(n - 2)(n + 2)



..which must be divisible by five since it is a
product of five consecutive integers. Thus,

3n^5 + 5n^3 + 7n == 0 (mod 5)

..from which we can conclude...

3n^5 + 5n^3 + 7n == 0 (mod 15).

Regards,
Kyle Czarnecki

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