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