By-Induction-Prove-that-12-n-4-n-2-

Question Number 2857 by Syaka last updated on 29/Nov/15

B y I n d u c t i o n P r o v e t h a t :

12 ∣ (n^{4} - n^{2})

Commented by Filup last updated on 29/Nov/15

What does ∣ mean ?

Commented by prakash jain last updated on 29/Nov/15

n^{4} - n^{2} is divisible by 12 .

Commented by Syaka last updated on 29/Nov/15

y e s, S i r

Answered by prakash jain last updated on 29/Nov/15

f (n) = n^{2} (n + 1) (n - 1) = (n - 1) n (n + 1) n

Statement 1 .

g (n) = (n - 1) n is divisible b y 2 . \dots (A)

g (1) = 0,

g (2) = 2

Let us say 2 ∣ g (n) for some n so n (n - 1) = 2 k

g (n + 1) = (n + 1) n = n^{2} + n = n^{2} - n + 2 n = 2 (k + n)

So the statement (A) is true \forall n .

It follows that h (n) = n (n + 1)

is divisible by 2 .

f (n) = g (n) h (n) is divisible by 4 . \dots (B)

Statement 2 :

y (n) = (n - 1) n (n + 1) is divisible b y 3 . \dots (C)

y (1) = 0

Let us y (n) is divisible b y 3 for some n .

y (n) = (n - 1) n (n + 1) = 3 j

y (n + 1) = n (n + 1) (n + 2) = n (n + 1) (n - 1 + 3)

= n (n + 1) (n - 1) + 3 n (n + 1) = 3 j + 3 n (n + 1)

= 3 (j + n^{2} + n)

H ence y (n) is divisible by 3 for all n .

From statement (B) and (C)

f (n) = (n - 1) n (n + 1) n is divisible 4 \times 3 = 12

since 3 and 4 {don}^{'} t have a common factor .

Commented by RasheedAhmad last updated on 29/Nov/15

6 t h l i n e : n (n - 1) = 2 k

7 t h l i n e : n^{2} - n + 2 n = 2 (k + 1) ?

n^{2} - n + 2 n = 2 k + 2 n = 2 (k + n)

Commented by prakash jain last updated on 29/Nov/15

Thanks Rasheed .

I have corrected the mistake .

Commented by Rasheed Soomro last updated on 29/Nov/15

g (n + 1) = (n + 1) n = n^{2} + n = n^{2} - n + 2 n = 2 (k + n)

So the statement (A) is true \forall n .

H o w i s 2 (k + n) o f t h e t y p e (n - 1) n

g (n) = (n - 1) n

g (n + 1) = 2 (k + n)

O n l h s n + 1 i n d i c a t e s t h a t y o u a r e s e e k i n g

r e s u l t f o r n + 1, b u t o n r h s y o u r e a c h e d a t 2 (k + n)

[i n s t e a d o f (\overset{―}{n + 1} - 1) (n + 1)]

W h y So the statement (A) is true \forall n . ?

Commented by prakash jain last updated on 29/Nov/15

g (n) = n (n - 1) = n^{2} - n

g (1) = 0 s o g (1) i s d i v i s i b l e b y 2 .

a s s u m e i f g (n) i s d i v i s i b l e b y 2 . g (n) = 2 k

g (n + 1) = (n + 1) (n + 1 - 1) = (n + 1) n

= n^{2} + n = n^{2} - n + 2 n = g (n - 1) + 2 n = 2 k + 2 n

= 2 (k + n) \Rightarrow 2 ∣ g (n + 1)

s i n c e 2 ∣ g (1) i s t r u e .

i f 2 ∣ g (n) \Rightarrow 2 ∣ g (n + 1)

h e n c e g (n) i s d i v i s i b l e b y 2 f o r a l l n \in N .

Commented by Rasheed Soomro last updated on 30/Nov/15

T h^{a} N k S S i r! I u n d e r s t o o d n o w .

Commented by Rasheed Soomro last updated on 30/Nov/15

EXCELLENT A p p r o a c h!