S-i-1-x-i-1-2-x-i-1-x-i-2-x-i-Z-Prove-that-S-Z-

Question Number 4645 by FilupSmith last updated on 18/Feb/16

S = \underset{i = 1}{\prod^{x}} (i^{\frac{1}{2} (x - i + 1) (x - i + 2)}), (x, i) \in Z

Prove that S \in Z

Answered by Yozzii last updated on 18/Feb/16

O b s e r v e t h a t 0.5 (x - i + 1) (x - i + 2) = \underset{j = 1}{\sum^{x - i + 1}} j

w h i c h i s a n i n t e g e r w h e r e w e a s s u m e t h e i n d e x o f t h e

s u m m a t i o n i s 0 < 1 ⩽ j ⩽ x - i + 1

s o x - i + 1 > 0 \Rightarrow x > i - 1 ⩾ 0 \Rightarrow x, i \in Z^{+} .

T h e r e f o r e, i^{\underset{j = 1}{\sum^{x - i + 1}} j} \in Z^{+} \Rightarrow S \in Z^{+} i f x, i \in Z^{+} .

Commented by FilupSmith last updated on 18/Feb/16

A way I thought to solve this was that :

a = (x - i + 1)

b = (x - i + 2) \Rightarrow b = (a + 1)

∴ S = i^{\frac{1}{2} a b}

a b = even

∵ either a or b is even while the other

is odd .

∴ a b = 2 n

∴ S = \sqrt{i^{2 n}}

∴ S \in Z

∴ \underset{i = 1}{\sum^{x - i + 1}} i^{\frac{1}{2} (x - i + 1) (x - i + 2)} is a sum of integers

in form :

\underset{i = 1}{\sum^{x - i + 1}} i^{n}

Is this an accurate proof ?

Commented by Yozzii last updated on 19/Feb/16

T h e f i r s t p a r t i s n t l e g i t, s i n c e y o u l e f t

o u t t h e Π f u n c t i o n i n t h e e q u i v a l e n t

e x p r e s s i o n f o r S . T h e k e y w a s y o u r

r e a l i s a t i o n o f i^{0.5 a b} a l w a y s b e i m g

a n i n t e g e r i f a ⩾ 1, b ⩾ 1 (b = a + 1) . S o t h e p r o d u c t i s a n i n t e g e r .

H e n c e S i s a n i n t e g e r .

I f a b < 0 a s i v a r i e s y o u h a v e t h e i^{0.5 a b} = \frac{1}{i^{- 0.5 a b}}

s o m o s t t e r m s i n t h e p r o d u c t w i l l b e

f r a c t i o n s . B u t I {d o n}^{'} t t h i n k y o u

m e a n t f o r x \in Z^{-} p o s s i b l y s o t h a t

1 ⩽ i ⩽ x \Rightarrow a b > 0 .

Commented by FilupSmith last updated on 19/Feb/16

I understand, thank you!