simple number theory ....

By binomial expansion -

3 ^{2 n} + 7

= ( 8 + 1 ) ⁿ + 7

= 1 + ⁿ C ₁ 8 + ............( multiples of 8 ) + 7

= 8 + ................multiples of 8

= multiples of 8

my sol : 3²ⁿ+7

= (3²)ⁿ -1+8

now 9ⁿ-1ⁿ is divisible by (9-1) or 8

therefore 3²ⁿ+7 is a multiple of 8

if n is even show that n(n²+20 ) is divisible by 48

n is what here minch? I thought of proving it by induction..if you put n = 1 you get 21 which is not divisible by 48.

The first one can be done by induction also.

For n = 1, we see that 3² + 7 = 16 is divisible by 8.

Let for n = k, 3^2k + 7 be a multiple of 8.
So 3^2k + 7 = 8@ for some natural @.

Now we must prove for n = k + 1.

3^{2(k + 1)} + 7
= 3^2k.3² + 7
= 9(8@ - 7) + 7
= 72@ - 63 + 7
= 72@ - 56
= 8(9@ - 7)

Hence for n = k + 1, the expression is a multiple of 8.
So by Principle of Induction, this holds for all natural numbers.

Okay, if n is even, substitute n by n = 2@, for some natural @(as n is always even, this condition holds good).

So our expression is 2@(4@² + 20).

For @ = 1, we see that
2(4 + 20)
= 48 is divisible by 48.

Now let for @ = k, our expression be divisible by 48.

2k(4k² + 20)
= 8k³ + 40k is divisible by 48.

So let 8k³ + 40k = 48X, for some natural X.

We must prove for @ = k + 1.

2(k + 1)((2(k + 1))² + 20)

I tried quite a lot to somehow get 8k³ + 40k into play...but it's not coming right. Can you check if I've done something wrong?

As DiPrit-H said, take n=2m

Let P(n)= 2m(4m²+20) is divisible by 48.
P(1) is true
We assume P(k)=2k(4k²+20) to be true
Then we have to prove P(k+1) is true
P(k+1)
=2(k+1)[4(k+1)²+20]
=2(k+1)[4k²+8k+4+20]
=2k[4k²+20] +2k(8k+4) + 2[4k²+8k+24]
=48J + 16k²+8k+8k²+16k+48 ......[48J=2k(4k²+20) which is divisible by 48]
=48J + 24k²+24k+48
=48J + 24[k²+k+2]

Now k²+k+2 is always even so 24[k²+k+2] is divisible by 48.

So P(k+1) is true
So, P(n) i.e. 2m(4m²+20]=n(n²+20) is divisible by 48

solution without induction
since n is even taking n=2k
now it becomes 2k(4k²+20) =8k(k²+5)
inference 1st: term divisible by 8

now taking two cases
a)k is even then k of k(k²+5) term is divisible by 2.
b)k is odd then (k²+5) of k(k²+5) is divisible by 2.
inference 2nd: term divisible by 2*8=16

now writing it as k(k²-1+6) it is divisible by 3(fermat's theorem)
inference 3rd: term divisible by 16*3=48.,

hence term is divisible by 48.

another method......
n(n²+20)=n(n² - 4 + 24) = n[(n-2)(n+2)+24]
=[(n-2)n(n+2)+24n]

Now (n - 2) n (n + 2) is the product of three consecutive even integers and
therefore must be divisible by 2 . 4 . 6 or 48 ; also 24n is divisible by 48

therefore n(n²+20) is divisible by 48.

induction gives the coolest solution,.,.,
observe for n=1,2,3......the identity holds good now...
3^2m+7 is divisible by 8 we r to show 3^(2m+2)+7 is also
3^(2m+2)=9.3^2m=8.3^2m+3^2m
hence proved./././.