Hi,

I have spent hours on this question:-.... no clue on how to tackle it. I appreciate if someonecould help.For what positive integers n is 4^{n}+ 7^{2n+1}+ 10^{3n-1 }aprime number?

Printable View

- June 6th 2012, 08:04 AMsnsengmodular arithmetic and prime number
Hi,

I have spent hours on this question:-.... no clue on how to tackle it. I appreciate if someonecould help.*For what positive integers n is 4*^{n}+ 7^{2n+1}+ 10^{3n-1 }aprime number?

- June 6th 2012, 08:33 AMSarasijRe: modular arithmetic and prime number
I claim that the sum S = 4

^{n}+ 7^{2n+1}+ 10^{3n-1}is always divisible by 3 for any n and is not a prime for all natural number n.

Proof:

4^{n}is of the form 3x + 1 i.e. 4^{n}= 3x + 1

7^{2n+1}= (3x2 + 1)^{2n+1}= 3y + 1 &

10^{3n-1}= (3x3 + 1)^{3n-1}= 3z + 1

So S = (3x+1) + (3y+1) + (3z+1) = 3(x + y + z + 1) => 3|S.

Hence proved. - June 6th 2012, 09:12 AMsnsengRe: modular arithmetic and prime number
Hi Sarasij,

Thanks so much for your speedy response.

In fact, I got the similar outcome that the expression is dividable by 3. This question was given by the school to my kid. I have spent a morning to learn from scratch about Modular Arithmetic and not much confidence on it. At the same time, I do not expect such a 'naughty' question as I presumed 'n' shall exist some positive integers... keep on trying. Ha Ha, interesting!

Have a good night/day. - June 6th 2012, 12:59 PMDevenoRe: modular arithmetic and prime number
modular arithmetic is the perfect tool for investigating things concerning divisibilty (and thus about primes, for example).

the nice thing about it is that:

a (mod n) + b (mod n) = a+b (mod n), and:

(a mod n)(b mod n) = ab (mod n)

so it doesn't matter if you reduce mod n before or after adding (or multiplying).

so we have:

4 = 1 (mod 3)

7 = 1 (mod 3)

10 = 1 (mod 3)

therefore:

4^{n}+ 7^{2n+1}+ 10^{3n-1}

≡ (1)^{n}+ (1)^{2n+1}+ (1)^{3n-1}(mod 3)

≡ 1 + 1 + 1 (mod 3)

≡ 0 (mod 3), thus we have 4^{n}+ 7^{2n+1}+ 10^{3n-1}is divisible by 3.

one can also attempt an induction proof, that 4^{n}+ 7^{2n+1}+ 10^{3n-1}is divisible by 3 for all natural numbers n.

base case: n = 1:

4 + 7^{3}+ 10^{2}= 4 + 343 + 100 = 447 = 3*149.

assume that:

4^{k}+ 7^{2k+1}+ 10^{3k-1}= 3x, for some positive integer x.

then:

4^{k+1}+ 7^{2k+3}+ 10^{3k+2}=

(4^{k+1}- 4^{k}) + (7^{2k+3}- 7^{2k+1}) + (10^{3k+2}- 10^{3k-1}) + 4^{k}+ 7^{2k+1}+ 10^{3k-1}

= 4^{k}(4 - 1) + 7^{2k+1}(7^{2}- 1) + 10^{3k-1}(10^{3}- 1) + 3x <---by our induction hypothesis

= 4^{k}(3) + 7^{2k+1}(48) + 10^{3k-1}(999) + 3x

= 3[4^{k}+ (7^{2k+1})(16) + (10^{3k-1})(333)] + 3x

= 3y + 3x = 3(y + x), for y = 4^{k}+ (7^{2k+1})(16) + (10^{3k-1})(333)

and is therefore true for all n. - June 6th 2012, 09:13 PMSarasijRe: modular arithmetic and prime number

Hi Devono,

Actually I don't know how to write the congruent modulo sign here. So that's why I explained the logic to him using only division algorithm. But it becomes much more simpler if I use this operator(congruent modulo). Can you please tell me how to use it in this forum?

I also admire your inductive trick based proof. (Bow) - June 7th 2012, 06:20 PMDevenoRe: modular arithmetic and prime number
there are two ways:

1. Use a unicode-based symbol (which you could do by copying and pasting this: ≡)

2. use latex (on this forum the latex tag is [tex ][/tex] (but without the space). in latex, the code for the modulus congruence sign is: \equiv such as:

what is inside the latex tags is:

2 \equiv 6\ \text{mod }4 - June 7th 2012, 11:25 PMSarasijRe: modular arithmetic and prime number