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;
β£zβ£=pβ, and
there exists a triangle whose three side lengths are p, the real part of z3, and the imaginary part of z3.
Solution:
Answer (349):
Assume that an odd prime p satisfies the conditions of the problem. Let c and d be integers with z=c+di, a=β£β£β£βRe(z3)β£β£β£β, and b=β£β£β£βIm(z3)β£β£β£β. Because c2+d2=β£zβ£2=p, it follows that a2+b2=p3. Note that (c+di)3=(c3β3cd2)+(3c2dβd3)i. Hence a=β£β£β£βc3β3cd2β£β£β£β and b=β£β£β£βd3β3dc2β£β£β£β. Let x=β£cβ£ and y=β£dβ£. Because x2+y2=p, it follows that x and y are relatively prime positive integers of different parity. Without loss of generality, assume that x>y. Then {a,b}={β£β£β£βx3β3xy2β£β£β£β,3x2yβy3}.
Now assume that a,b, and p, with a>b, are the sides of the triangle that satisfies the conditions of the problem. Because 2a2>p3, it follows that a is the longest side of the triangle, and hence a,b, and p form the sides of the triangle if and only if β£aβbβ£<p.
If xβ€y3β, then the Triangle Inequality condition is equivalent to
which simplifies to x(1βΞ±)<y(1+Ξ±). This implies
x<1βΞ±1+Ξ±βy=3ββ13β3ββy=3βy,
which is false. Therefore β£β£β£βxβΞ±yββ£β£β£β<1.
Because x and y are of different parity, for each positive integer y, there is at most one positive integer x such that x and y lead to the desired triangle with p=x2+y2, namely x is either xcβ or xcβ+1, where xcβ=β(2+3β)yβ.
Note that if yβ₯9, then x>3.6β y>32, leading to values for p greater than 92+322>1000. Checking potential values of y from 8 down for prime values of p=x2+y2 leads to the following table:
y
x
xΒ² + yΒ²
Prime?
8
29
905
NO
7
26
725
NO
6
23
565
NO
5
18
349
YES
4
15
241
YES
3
12
153
NO
2
7
53
YES
1
4
17
YES
If x=5 and y=18, then the Triangle Inequality condition is (5+18)(4β 5β 18β52β182)<52+182, which is equivalent to 23β 11<349, which is true. Furthermore, 52+182=349 is a prime number. The triangle in question has sides
(x3β3xy2,3x2yβy3,p)=(4482,4735,349).
The requested prime number is 349 .
OR
The answer can be obtained through approximation. For example, let z=x+iy=pβ(cosΞΈ+isinΞΈ) so that z3=ppβ(cos(3ΞΈ)+isin(3ΞΈ)). Because the real and imaginary parts of z3 must be positive, it follows that 0<ΞΈ<6Οβ. By the triangle inequality
β£ppβcos(3ΞΈ)βppβsin(3ΞΈ)β£<p,
from which
sin(6ΞΈ)>1βp21β,
so ΞΈβ12Οβ. Then xβcot(12Οβ)y=(2+3β)y. This can be used to build the table from the first solution using small positive integer values of y, and the solution follows.