Question

$\bf{Find\: all\: prime\: numbers\: p\: and\: q\: such\: that:}$
$\boxed{p+q=(p-q)^3}$
Justify your answer and explain.
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Answer 1

p3−q3=p⋅q3−1⟺p3+1=p⋅q3+q3⟺p3+1=q3⋅(p+1)⟺p3+1p+1=q3⟺p2−p+1=q3p3−q3=p⋅q3−1⟺p3+1=p⋅q3+q3⟺p3+1=q3⋅(p+1)⟺p3+1p+1=q3⟺p2−p+1=q3

Note that p+1≠0p+1≠0 because −1−1 is not a prime number and that

p3+1=(p2−p+1)⋅(p+1).p3+1=(p2−p+1)⋅(p+1).

So we have to find all pairs (p,q)(p,q) of prime numbers such that p2−p+1=q3.p2−p+1=q3.
 p2−p+1=q3p2−p+1=q3

As proved here above the unique pair is (19,7).(19,7).

(We can verify it : 193−73=19⋅73−1=6516.)193−73=19⋅73−1=6516.)

Answer 2

How is this possible