I need some advice on how to proceed on with this question:
Determine all possible quadruples of positive integers $a,b,c,d\in\mathbb{Z}_{>0}$ such that
$a^3=b^2$ , $c^3=d^2$ and $c-a=7$.
So far I have gathered the following:
$c^3-a^3=d^2-b^2$ and so $(c-a)(c^2+ac+a^2)=(d-b)(d+b)$
$7((c-a)^2+3ac)=(d-b)(d+b)$
$7(49+3a(a+7)=(d-b)(d+b)$
$7(3a^2+21a+49)=(d-b)(d+b)$
and I'm stuck here. However I do realised that $(3a^2+21a+49)$ is unfactorizable. So does it imply that $(3a^2+21a+49)$ is prime too?
If so I can conclude that (d-b)(d+b) is a product of 7 and a prime number. To which, either
$(d-b)=7, (d+b) = $ a prime number OR $d-b= $ a prime number, $d+b=7$
Would greatly appreciated if someone can enlighten me on this.
Thank you very much!
To answer your question: there is no known formula to generate primes only, so no.
Now, since $a^3=b^2$ and $c^3=d^2,$ the Unique factorization Theorem implies that the exponent of any prime factor of $a$ must be even and the same happens with $c$ and so $c-a$ is a difference of squares and hence...