Show that 2 is the only prime which is the sum of 2 positive cubes?

Question

Any info or guidelines to help solve would be appreciated. I'm sure it has to do with one cube being odd and the other even and any prime larger than 2 has to be odd.

Answer 1

math_kp has the correct answer, which turns out not to have much to do with primes larger than 2 being odd. I'm finding that answer a little hard to follow, though, so I think it deserves a rewrite. Here's my try at explaining it.

Suppose that some number n is expressible as the sum of the cubes of two positive integers, a and b, where

a ≥ b

(because we can arbitrarily choose which of them is represented by which variable).

Now,

n = a³ + b³ = (a + b) (a² - ab + b²)

and it is clear that (a + b) is greater than 1 because a and b are both positive integers.

(a² - ab + b²) = a (a - b) + b²

which is plainly larger than 1 unless a=b and b=1 (because we specified that a would represent the larger, if any, of a and b).

If a=b=1, n = 2.

If not, n has two factors greater than 1 and is not a prime.

Q.E.D.

Answer 2

well.....1^3 + 1^3 = 2

and 2 is a prime 'cause it can only be divided by itself and by 1

...now how do you show that it is the ONLY prime?

I don't think any other cubed numbers add to a prime number....but I can't think why or how to prove it right now.....I hope I gave something to start with

Answer 3

The way your question is phrased, the cubes do not have to be the same.

Then 2^3 = 8 and 5^3 =125 will sum to 133 and that is prime.

If the cubes have to be the same, then we can deduce a proof.

Any odd number when cubed will also be an odd number.  Sum 2 odd numbers and it is even, thus it is not prime.  Even numbers cubed are even and also not prime.

Answer 4

Well ..I don't thnk it is possible becoz ... cubes of 2 +ve numbers are always +ve ..So ..It can't result in anythng less than 2 .

   It may be possible if ..we choose the number as 1 .

So ..1^3+1^3 = 2.

  So..I think it is possible by Hit & Trial method ..

Answer 5

a^3 + b^3 = (a+b)(a^2-ab+b^2)

we know a+b is > 1 (as a>= 1 and b >= 1)



if  = a = b and b > 1 then a^2+ b^2 - ab = b^2 > 1

without loss of generality we can choose the larger number a

if a > b then a^2 + b^2 - ab = a (a-b) + b^2 > b^2 that is > 1

as 2 factors > 1 so it cannot be prime

proved math kp

Answer 6

Let P = a prime which is equal to the sum of two positive cubes a^3.

Then:

a^3 + a^3 = P

or 2 (a^3) = P

Therefore P is a multiple of 2

Therefore P is even.

There is only one even prime, and that is 2.  All other primes are odd.

Answer 7

1. Show that every prime number other than 2 is an odd number. This is so because every prime number larger than 3 would be divisible by 2 if it was even.

2. Show that by adding any two odd or even numbers you always get an even number. You HAVE to add one even number to an odd number to recieve an odd number.

3. No two of the same number added together can be odd, and all prime numbers greater than 2 are odd.