# for PH and others so inclined

• Dec 22nd 2006, 02:07 PM
galactus
for PH and others so inclined
"Show that $\displaystyle \frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}$

is always an integer for integral values of x".
• Dec 22nd 2006, 02:22 PM
Plato
It is relatively easy (but messy) to show that for each positive integer N
$\displaystyle 3N^5 + 5N^3 + 7N$ is a multiple of 15.
an other
• Dec 22nd 2006, 02:43 PM
galactus
That was a fast response, Plato;) . I noticed, since the semesters have come to a close, there hasn't been much posting.

Anyway, maybe I am off base, but $\displaystyle \frac{x^{5}}{5}+\frac{x^{3}}{3}+\frac{7x}{15}=\fra c{x^{5}-5}{5}+\frac{x^{3}-3}{3}+x$

Now, use Fermat's theorem:

$\displaystyle x^{5}-x\equiv{0} \;\ mod(5)$

and

$\displaystyle x^{3}-x\equiv{0} \;\ mod (3)$

Therefore, it's an integer.

Seems rather logical. But, I am not that much of a number theorist. Have been learning it, though.
• Dec 23rd 2006, 12:54 AM
earboth
Quote:

Originally Posted by Plato
It is relatively easy (but messy) to show that for each positive integer N
$\displaystyle 3N^5 + 5N^3 + 7N$ is a multiple of 15.
an other

Hello Plato,

use mathematical induction to prove this property:

Prove that

$\displaystyle 3n^5+5n^3+7n=15 \cdot k\ , \ k \in \mathbb{Z}$

$\displaystyle step\ 1:\ n=1\ :\ 3+5+7=15=15 \cdot 1 \text{ is true}$

$\displaystyle step \ 2:\ assumption: 3n^5+5n^3+7n=15 \cdot k \text{ is true}$

$\displaystyle step\ 3:\ conclusion: 3(n+1)^5+5(n+1)^3+7(n+1)=$

$\displaystyle 3n^5+15n^4+30n^3+30n^2+15n+3+5n^3+15n^2+15n+5+7n+7 =$

$\displaystyle \underbrace{3n^5+5n^3+7n}_{k\cdot 15}+15(n^4+2n^3+3n^2+2n+1)+3+5+7=$

The 1rst summand is a multiple of 15 according assumption, the 2nd summand is a multiple of 15 because it contains the factor 15 and the last 3 summands add up to 15. Thus $\displaystyle 3(n+1)^5+5(n+1)^3+7(n+1)=$ is multiple of 15 and therefore the asumption is true for all $\displaystyle n \in \mathbb{N}$

EB

Merry Christmas and a happy New Year.