# Prove that 14^n+11 is never prime

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username53983805

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#1

I'm trying to use an inductive argument, however I've gotten stuck at one part.

So for n=0, not a prime number.

Suppose that is not prime for some integer . Let for .

. We know q is not prime from above, and 13(q-11) cannot be prime as q-11=1 for this to be true, and if , however for , so q cannot equal 12. Therefore 13(q-11) is not prime for all q.

I'm up to this point, and then wanted to show that this value can never be prime, but showing that both q and 13(q-11) are never prime does not imply that their sum is never prime (8+9=17 but 8 and 9 are clearly not prime). So is there anyway I could carry on from here to show that their sum is never prime? My only thought is some sort of contradiction argument?

Edit: I think I have it. Suppose that is prime, so let for some prime p. as . However, as we know q is odd ( will always be even, and an even + odd yields an odd number), an odd number must divide the RHS. But the only way we have an odd number on the RHS is if p is even, so p must equal 2. However this clearly cannot be true, so p is a prime. RHS is even, and so we have a contradiction when we stated that q|RHS.

Hence is never prime, and it follows by induction this is true for all . I hope this is correct.

So for n=0, not a prime number.

Suppose that is not prime for some integer . Let for .

. We know q is not prime from above, and 13(q-11) cannot be prime as q-11=1 for this to be true, and if , however for , so q cannot equal 12. Therefore 13(q-11) is not prime for all q.

I'm up to this point, and then wanted to show that this value can never be prime, but showing that both q and 13(q-11) are never prime does not imply that their sum is never prime (8+9=17 but 8 and 9 are clearly not prime). So is there anyway I could carry on from here to show that their sum is never prime? My only thought is some sort of contradiction argument?

Edit: I think I have it. Suppose that is prime, so let for some prime p. as . However, as we know q is odd ( will always be even, and an even + odd yields an odd number), an odd number must divide the RHS. But the only way we have an odd number on the RHS is if p is even, so p must equal 2. However this clearly cannot be true, so p is a prime. RHS is even, and so we have a contradiction when we stated that q|RHS.

Hence is never prime, and it follows by induction this is true for all . I hope this is correct.

Last edited by username53983805; 9 months ago

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ghostwalker

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#2

(Original post by

I'm trying to use an inductive argument, however I've gotten stuck at one part.

So for n=0, not a prime number.

**username53983805**)I'm trying to use an inductive argument, however I've gotten stuck at one part.

So for n=0, not a prime number.

Edit: I think I have it. Suppose that is prime, so let for some prime p. as . However, as we know q is odd ( will always be even, and an even + odd yields an odd number), an odd number must divide the RHS. But the only way we have an odd number on the RHS is if p is even, so p must equal 2. However this clearly cannot be true, so p is a prime. RHS is even, and so we have a contradiction when we stated that q|RHS.

Hence is never prime, and it follows by induction this is true for all . I hope this is correct.

Hence is never prime, and it follows by induction this is true for all . I hope this is correct.

An inductive proof would work, but it's considerably more complicated.

As a hint, look for specific factors in the sequence

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#3

Can't edit my previous post

But once you have an inductive proof, or even without it, you should be able to rewrite it more simply without the necessity of induction.

But once you have an inductive proof, or even without it, you should be able to rewrite it more simply without the necessity of induction.

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#4

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Minor point: = 12

Just becase an odd number divides the RHS, does not imply the RHS is odd, indeed the 14 on the LHS imples the RHS must be even.

An inductive proof would work, but it's considerably more complicated.

As a hint, look for specific factors in the sequence

**ghostwalker**)Minor point: = 12

Just becase an odd number divides the RHS, does not imply the RHS is odd, indeed the 14 on the LHS imples the RHS must be even.

An inductive proof would work, but it's considerably more complicated.

As a hint, look for specific factors in the sequence

Why does the 14 on the LHS imply that the RHS

*must*be even?

From terms in the sequence, I'm getting the fact that when n is odd, the final digit is always a 5 so can never be prime, and when n is even, there is always a factor of 3, and so can never be prime either. I now have to generalise this right?

Last edited by username53983805; 9 months ago

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#5

(Original post by

i have no clue where I got 15 from haha.

Why does the 14 on the LHS imply that the RHS

**username53983805**)i have no clue where I got 15 from haha.

Why does the 14 on the LHS imply that the RHS

*must*be even?
From terms in the sequence, I'm getting the fact that when n is odd, the final digit is always a 5 so can never be prime, and when n is even, there is always a factor of 3.

Now think in terms of modular arithmetic - and forget the induction.

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#6

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Well 14 is even, 2|LHS, therefore 2|RHS

So, alternate terms are divisible by 3 and 5.

Now think in terms of modular arithmetic - and forget the induction.

**ghostwalker**)Well 14 is even, 2|LHS, therefore 2|RHS

So, alternate terms are divisible by 3 and 5.

Now think in terms of modular arithmetic - and forget the induction.

Okay let me have a think

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(Original post by

But could I not just argue that q|LHS so q|RHS ==> q|p and q|143

**username53983805**)But could I not just argue that q|LHS so q|RHS ==> q|p and q|143

**Edit: Should have said in previous post that consecutive terms are alternately divisible by 3 and 5, but am unable to edit it due to a TSR bug.**

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#8

**ghostwalker**)

Well 14 is even, 2|LHS, therefore 2|RHS

So, alternate terms are divisible by 3 and 5.

Now think in terms of modular arithmetic - and forget the induction.

For even n, i.e for , want to prove that .

for all . (I am not 100% sure about this though).

For odd n, i.e for , want to prove that .

for all .

Again I am not sure about the validity of these proofs

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#10

(Original post by

Think I've got it.

For even n, i.e for , want to prove that .

for all . (I am not 100% sure about this though).

For odd n, i.e for , want to prove that .

for all .

Again I am not sure about the validity of these proofs

**username53983805**)Think I've got it.

For even n, i.e for , want to prove that .

for all . (I am not 100% sure about this though).

For odd n, i.e for , want to prove that .

for all .

Again I am not sure about the validity of these proofs

And you can finish with a summary line of some description.

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#11

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They look fine. In your last line just decide whether you're going with "m" or "k".

And you can finish with a summary line of some description.

**ghostwalker**)They look fine. In your last line just decide whether you're going with "m" or "k".

And you can finish with a summary line of some description.

Out of curiosity, would there of been any way I could of proceeded in my original attempted solution to prove that was never prime, as that seemed to be all I needed to finish the proof.

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(Original post by

Out of curiosity, would there of been any way I could of proceeded in my original attempted solution to prove that was never prime, as that seemed to be all I needed to finish the proof.

**username53983805**)Out of curiosity, would there of been any way I could of proceeded in my original attempted solution to prove that was never prime, as that seemed to be all I needed to finish the proof.

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#13

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Don't know, but I have grave doubts. q being composite (and odd) is insufficient to prove is not prime, e.g. q=15, gives 67, a prime, in the formula. So, you'd need to use some other information you know or can deduce.

**ghostwalker**)Don't know, but I have grave doubts. q being composite (and odd) is insufficient to prove is not prime, e.g. q=15, gives 67, a prime, in the formula. So, you'd need to use some other information you know or can deduce.

Last edited by username53983805; 9 months ago

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