Two Different Two Squares

Thanks. I've updated the problem statement. Those who previously answered 25 will be marked correct.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “line line line” menu in the top right corner. This will notify the problem creator who can fix the issues.

Is there an approach other than hit and trial?

'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 $N$ .

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).' - Mark Henning

From this, we can easily cut down on the number of cases having to be checked. We get that 50 is the smallest number satisfying.