Algebra help please!?

Question

prove that the sum of two consecutive odd primes has at least three prime divisors

Answer 1

Nice one.

It's always going to be divisible by 2, obviously.

It will also be divisible by 3, since 3 divides neither of the primes themselves yet must divide exactly one of any set of three consecutive integers.

However, the statement is false. 11 and 13 are consecutive odd primes, and the only primes that divide 24 are 2 and 3.

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EDITS BASED ON ANSWERS BELOW:

Thanks for the nice words, Puggy!

Guys, 1 is NOT prime.

I was assuming that we were talking about consecutive odd numbers that both were prime. Otherwise, even divisibility by 3 may not follow; e.g., 31 + 37 = 68.

If the question is at least 3 prime factors COUNTING dupes, and we take the n/n+2 meaning, it's easy. We get two factors of 2 and one of 3 by the methods above.

Now the next question is -- for what values of j and k can (2^j)(3^k)+-1 both be prime? j=2, k=1 gets the job done, as do j=1, k=1 and j=1, k=2. For that matter, j=3, k=2 does too (72).

We're not on a winning streak here. When n ends in 2 or 8 ( (needed for a factor of 5 not to show up in n-1 or n+1), we're getting a number of prime pairs.

Answer 2

This statement is not true in all cases. Take 3 and 5. The sum is 8 with factors of 1, 2, 4, and 8.
Take 7 and 11 with sum of 18 and factors of 1, 2, 3, 6, 9, and 18. The first case has only 1 prime divisor and the second case only 2 prime divisors.

Answer 3

Let x be the odd prime number.
Since x is the odd prime number, the other odd prime number is x+2. They are two digits apart.

Sum of two consecutive prime number: (x) + (x+2) = 2x + 2
So, it will be divisible by 2 as seen from above, as well as 1.

To find the other one, try substituting some numbers into the equation.

Now, let's substitute 5 into the x.
So, 2(5) + 2 = 12

For 12, it will be divisible by the prime numbers 1,2,3.

Sum of different numbers have different divisor (that 1,2,3 above) so it depends but they will definitely be divisible by 1 and 2.

Answer 4

I'm not sure that is true, because one obvious example disputes it: 1 and 3 are consecutive odd primes, and their sum, 4 , hasn't 3 prime divisors, does it? Or am I missing something?

Answer 5

Let p and q be two consecutive odd primes.

Then, this means p and q can be expressed in the form 2_ + 1; that is there exists i and j such that
p = 2i + 1
q = 2j + 1

Added together,

p + q = 2i + 1 + 2j + 1
= 2i + 2j + 2

Note that we can factor out a 2;
= 2(i + j + 2)

And this alone demonstrates 2 to be a divisor of (p + q) {and 2 is prime}.

I admit I got stuck at this point.

Edit: Curt, you're a genius for actually testing it out instead of being an idiot like me who assumed the proof was true and struggled to prove it. :)

Answer 6

I like Curt's answer. Perhaps the original question was really that the sum has three prime factors (counting multiplicity)?

So 24 = 2^3 * 3, so it has 4 prime factors.

Answer 7

For increasing cubic binomials the favourite formula is as follows: (a + b) ^ 3 = a^3 + 3*a^2*b^a million + 3*a^a million*b^2 + b^3 on your case, a is x and b is -y^5 So (x - y^5)^3 = x^3 + 3*x^2*(-y^5)^a million + 3*x^a million*(-y^5)^2 + (-y^5)^3 Simplified: =x^3 - 3x^2*y^5 + 3x*y^10 - y^15 :D