Problem:
Find the largest prime number p<1000 for which there exists a complex number z satisfying
the real and imaginary part of z are both integers;
∣z∣=p, and
there exists a triangle whose three side lengths are p, the real part of z3, and the imaginary part of z3.
Solution:
Let's denote z=a+bi
After expanding z3, we'd have that
z3=(a+bi)3=a3−3ab2+(3a2b−b3)i
The only condition we are given is that a triangle exists with the real and imaginary parts of z3 and this prime p, so let's create inequalities using Triangle Inequality to start out and bound a and b
We have that
a3−3ab2<p+3a2b−b3→∣(a3−3ab2)−(3a2b−b3)∣<p
∣a3+b3−3ab2−3a2b∣<p
Since we have that ∣z∣=p, we have that ∣z∣=p2=a2+b2
So we can rearrange our original inequality to write
∣a3+b3−3ab2−3a2b∣<a2+b2
From here, we have to use some sort of rational guess and check.
Notice we can factor a3+b3−3ab2−3a2b=(a+b)(a2+b2−4ab)
So we'd have that
∣a+b∣∣a2+b2−4ab∣<a2+b2
We can assume WLOG that a>b, since i3(a+bi)=b+ai so we could transform to get a solution flipping a and b.
From here, let's also denote x=ba.
Notice also that the root of x2−4x+1 is 2±3 which is important because that would be a root to a2+b2−4ab if we divided off by b2 and rewrote the equation in terms of x.
Let's consider bounding x≥4, particularly as it is close to 2+3.
In particular, we'd have the equation
∣b−4b∣∣16b2−16b2+b2∣≤17b2→3b3≤17b2→b≤317
or for integers, b≤5
If b≥6, Then we'd have that a=23b and then consequently 23b<a<4b
Bounding this accordingly, 1000>p and p=a2+b2 so we'd have
1000>a2+b2≥(23b)2+b2≥13b2
or b≤8.
Testing out values for b=8 and b=7, we'd have no solution where a2+b2 is also prime.
But testing b=5, we get a valid solution, namely b=5, a=18 that yields 52+182=349, which is prime.
If b≤4, we'd have that
p>(a+b)(a2+b2−4ab)≥1144>1000
which wouldn't give us a valid solution so we are done.