Figure out all primes p and q such that Figure out all primes p and q such that
$p^3$ +  19$q^3$ + 2018 is the cube of a prime. 
 A: try HINT: Notice that one of $p$ and $q$ must be even. Otherwise, $r$ would be even. Try first $q=2$. Then we have $p^3+1246=r^3$, which yields $(r-p)(r^2+rp+p^2)=1246=2*7*89$.
Then the case where $p=2$.
A: Let r be the prime such that 
$p^3 + 19q^3 + 2018 = r^3$
Since 2 is the only even number prime but r does not equal to 2 (trivial), by observing the even-odd of the LHS and RHS, either $p=2$ or $q=2$ .
If $p = 2$, $2^3 + 19q^3 + 2018 = r^3$
Let $s^3 = 19q^3$
Since 19 is not a perfect cube, s or $s^2$ cannot be a rational number, hence $p=2$ cannot be a correct answer
If $q = 2$, $p^3 + 19*2^3 + 2018 = r^3$
$p^3+ 19*8 + 2018 = r^3$
$r^3 – p^3 = 2170$
$(r-p)(r^2+rp+p^2) = 2170$
2170 is not prime, but for the product of the 2 factors $(r-p)$ and $(r^2+rp+p^2)$ ends in 0 as last digit, one of the factor is divisible by 5. 
EITHER  
$(r – p)$ is divisible by 5………………….(A)
Since 5 is the only prime divisible by 5, and both r or p are odd numbers, $(r – p)$ must be divisible by 2 as well – hence, $(r – p)$ is divisible by both 5 and 2 – and is divisible by 10
let $r – p = 10k$ where k is an integer
$10k ((r^2+rp+p^2) = 2170$
$k(r^2+rp+p^2) = 217$ (a prime), so $k=217$ (incorrect) or
$r^2+rp+p^2 = 217$ and $k=1$
since  $√217 = 14.73$, we have $r,p<14$
By trial and error we get $(r,p) = (3,13)$ or $(13,3)$ – 
but since $r-p = 10$, $r=13, p=3$ is the correct answer
OR
$(r^2+rp+p^2)$  is divisible by 5………………..(B)
By checking the last digit of $r^2+rp+p^2$ when p and q are prime numbers not equal to 5, the last digit can only be 1,3,7,9 for all combinations of non-2 primes (r,p).  The only way $r^2+rp+p^2$ can be divisible by 5 is unless (r,p) = (5,5) which is not a correct answer. So there is no solution in (B)
Therefore, $(p,q,r) = (3,2,13)$
Check – $3^3 + 19 * 2^3 + 2018 = 2197 = 13^3$
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