2 Prime Numbers Problem

p+q= (p-q)^2 * (p-q), then, (p - q) divides (p + q). This implies that (p - q) divides 2q.

If (p - q) divides q, then (p - q) = 1 or (p - q) = q, which are impossible statements. Now, notice that (p - q) divides 2. As (p - q)=1 is impossible for prime numbers, then (p - q) = 2.

From the original equality, we have p + q = 8, then the only prime numbers that satisfy this are 5 and 3.

From (p - q)^3 >0 we know that p>q, then p = 5 q = 3

Thus p - q + 2pq = 5 - 3 + 2(5)(3) = 32

Here's a solution that I don't think anybody has mentioned yet.

Suppose that $p,q>3$ . Note that all primes $p$ greater than $3$ satisfy $p\equiv \pm 1\pmod 6$ . This means that the residues of $p$ and $q$ are either equal to each other or negatives of each other. But then exactly one of $p+q$ or $p-q$ is divisible by six, contradiction! (For example, if $p\equiv q\equiv 1\pmod 6$ , then $p-q\equiv 0\pmod 6$ while $p+q\equiv 2\pmod 6$ . But it must hold true that $p+q\equiv (p-q)^3\equiv 0^2\equiv 0\pmod 6$ , which doesn't hold.) Therefore one of $p$ or $q$ must be two or three. After some experimentation, we find $(p,q)=(5,3)$ does indeed work, and the requested expression has value $5-3+2\cdot 5\cdot 3=\boxed{32}$ .

the prime no.s are 5 and 3. LHS: 5+3=8. RHS: (5-3)^3=(2)^3=8. so p-q+2pq=5-3+30=32

let (p-q)=n then we have p-q=n and p+q=n^3 therefore we have ; p=n(n^2+1)/2 and q= n(n^2-1)/2; since p and q have to be prime they should get divided only by themselves or 1 therefore this leaves us with the choice that n has to be 2. Therefore p=5 and q=3. and so answer is 32.

(p-q)^3=p+q <=> 2q=(p-q-1)(p-q)(p-q+1) ;p>q>=2 SO p-q-1=1 and p-q=2 and p-q+1=q =>p=5,q=3 =>p-q+2pq=32

it can be easily proved that p or q cannot be equal to 2.

The possible value of p+q: 8,27,64,125, etc. //(p-q)power3

p-q is even. So their cube is also even.p+q:8,64,216,etc.

if the sum of two number is 8,64,216,etc, the minimum value of p-q+2pq is 8,64,216,etc.(simple maths,its same as sum only for this expressionexpression)

Given all options are less than 64..so sum must be equal to 8. Now from simple counting,it can be shown that the numbers are 5 and 3 and the correct solution is 32

Easily check for the smallest cube after $1$ that is $8$ . It can be written as $5 + 3$ where $5$ and $3$ are prime. Now $5 + 3 = {(5 - 3)}^3 = 8$ . So, $p - q + 2pq = 5 - 3 + 2×5×3 = \color{#69047E}{\boxed{32}}$ .

(5+3)=8 and (5-3)^3=8 then p=5 , Q=3; so 5-3+2 * 5 * 3=32

p-q divides p+q likewise If (p - q) divides q, then (p - q) = 1 or (p - q) = q, which are impossible statements. Now, notice that (p - q) divides 2. As (p - q)=1 is impossible for prime numbers, then (p - q) = 2 so we will get p=5 q=3 by solving it we get 32 as answer .

Hello,

stated that p and q are prime numbers = [ 1,2,3,5,7,11,......]

p + q = (p - q)^3

(p + q)^(1/3) = (p-q)

p > q , i picked p = 5 , q = 3

so (5 + 3 )^(1/3) = ( 5 - 3 ) = 2,

therefore p - q + 2pq = 5 - 3 + 2(5)(3) = 32...

thanks ah....

since p+q/p-q=(p-q)^2 by putting p=3&q=2 then, LHS is not equal to RHS then we go for next two prime number i.e.5 and 3 and we get LHS= RHS.By putting this we get 32 as required answer.

Let the prime numbers be 5 and 3 for p and q respectively, we know that p>q since (p-q)>0, for the first equation we have 5+3=(5-3)^3=8. Thus, for the second equation we'll have 5-3+2 5 3=32. Which is the right answer.

first you have to think what is the value of (p-q)^3. if p-q=1 then (p-q)^3=1. it doesn't satisfy any p,q prime value. if p-q=2, then (p-q)^3=8. on left side p=5,q=3, so p+q=8, right side, (5-3)^3=8. So p=5, q=3 are the solutions. now just put them on the equation to determine the desired value.