Not A Taxicab Number

Relevant wiki: Gaussian Integers

The primes in the Gaussian integers are:

• prime integers congruent to $3$ modulo $4$ ,
• numbers of the form $a + bi$ where $a^2 + b^2$ is a prime integer congruent to $1$ modulo $4$ ,
• $1 + i$

and their associates. An integer $N$ can be written as the sum $X^2 + Y^2$ of two positive integers if $N$ can be written as $(X+iY)(X-iY)$ in the Gaussian integers. Since $2 = (1+i)(1-i)$ , any power of $2$ can be so written. Similarly, any power of a prime integer congruent to $1$ modulo $4$ can be so written, but prime integers congruent to $3$ modulo $4$ cannot. Thus $N$ can only be written as a sum of two squares when prime integers congruent to $3$ modulo $4$ occur an even number of times in the prime decomposition of $N$ . There are $30$ numbers between $2000$ and $2099$ inclusive with this property.

Since $1+i$ and $1-i$ are associate, $(1+i)^m$ divides $X+iY$ precisely when $(1-i)^m$ divides $X-iY$ . Thus, if $2^m$ is a factor of $N$ , we must have $(1+i)^m$ as a factor of $X+iY$ , so powers of $2$ do not give us any flexibility in the choice of $X+iY$ .

To be able to write $N$ as a sum of two squares in at least two different ways, $N$ must contain at least 2 prime factors which are congruent to $1$ modulo $4$ (these prime factors could be repeated). There are $9$ numbers which satisfy all conditions so far: $2000 \,,\, 2005\,,\, 2020\,,\, 2025\,,\,2041\,,\,2045\,,\,2050\,,\,2074\,,\,2080$ Of these, all but two have two different odd prime factors, and so certainly can be written as a sum of square positive integers in least two different ways. The remaining two cases are:

• $2000 \,=\, 2^4 \times 5^3$ . We can have $X+iY = 4(2+i)^3$ and $X+iY = 20(2+i)$ , so this number works.
• $2025 \,=\, 3^4 \times 5^2$ . We are forced (to within association) to have $X + iY = 9(2+i)^2$ , so $2025$ can only be written as a sum of two square positive integers in one way (the solution $X + iY = 9(2+i)(2-i)$ is not possible, since the option of $45^2+ 0^2$ is not allowed).

Thus there are $\boxed{8}$ possible numbers.

A brute force solution (python):

sumlist = []
for a in range(1,2100):
    for b in range(1,2100):
        if 2000<=a**2+b**2<=2099:
            sumlist.append(a**2+b**2)
a=[]
for i in sumlist:
    if sumlist.count(i)>=4:
        a.append(i)
print (len(set(a)))