C L's number

First we prove that $M$ only has prime divisor of the form $4m+1$ . Assume that there exists a prime $p=4k+3$ divides $M=a^2+b^2$ . From the given condition, we know that if $a$ or $b$ is divisible by $p$ , it will result in $M$ is divisible by $p^2$ , a contradiction, so $gcd(a;p)=gcd(b;p)=1$ . We have: $a^2 \equiv -b^2 (mod p)$ $\Leftrightarrow a^{4k+2} \equiv -b^{4k+2} (mod p)$ By Fermat's Little Theorem, it is equivalent to $1 \equiv -1 (mod p)$ , a contradiction. So any prime divisor of $M$ must have the form $4m+1$ . We have $M=p_1p_2…p_k$ , where $p_i$ are distinct primes of the form $4m+1$ . We shall prove that $k \geq 3$ .

For $k=1$ , since $M=p_1$ is a prime, according to the famous Fermat-Euler’s theorem, $M$ can only be expressed uniquely as the sum of 2 squares, so $M$ does not satisfy the given conditions.

For the case $k=2$ , we have $M=p_1p_2$ , and because of Fermet-Euler’s theorem, we can write $p_1=x^2+y^2; p_2=u^2+v^2 (x>y; u>v)$ . Now assume that $M$ can be expressed as the sum of two squares for more than 2 ways, that is $M=a^2+b^2=c^2+d^2=e^2+f^2$ . We may suppose that $a>c>e>f>d>b$ . We have:
$a^2 \equiv -b^2 (mod M)$ $c^2 \equiv -d^2 (mod M)$

Hence $a^2c^2 \equiv b^2d^2 (mod M) \Leftrightarrow (ac-bd)(ac+bd)$ is divisible by $M$ .

Since $M^2 =(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2 \geq (ac-bd)^2$ , we cannot have $M|(ac-bd)$ or $M|(ac+bd)$ , therefore exactly one of the two numbers $(ac-bd),(ac+bd)$ is divisible by $p_1$ , and the other number is divisible by $p_2$ . Similarly we can deduce the same property for 2 numbers $(ad-bc),(ad+bc)$ . Moreover, if both $(ad+bc)$ and $(ac+bd)$ is divisible by $p_1$ (or $p_2$ ), it will result in $a^2 \equiv b^2 \equiv -b^2 (mod p_1)$ , a contradiction. So WLOG we may assume that $p_1|(ad+bc)$ and $p_2|(ac+bd)$ , which also means that $p_1|(ac-bd)$ and $p_2|(ad-bc)$ . Write $ac+bd=mp_2; |ad-bc|=np_2$ , we have $p_1^2p_2^2=(ac+bd)^2+(ad-bc)^2=p_2^2(m^2+n^2) \Rightarrow m^2+n^2=p_1^2$ . This is Pythagorean equation, and since $gcd(m;n)=1$ and $p_1$ can only be uniquely expressed as $x^2+y^2$ , we must have { $m;n$ }={ $x^2-y^2;2xy$ }.

Hence { $ac+bd;|ad-bc|$ } = { $(x^2-y^2)p_2;2xyp_2$ }. Similarly, we can prove that { $ad+bc;ac-bd$ } = { $(u^2-v^2)p_1;2uvp_1$ }. So { $ad+bc;ac+bd;|ad-bc|;ac-bd$ } = { $( x^2-y^2)p_2;2xyp_2;2uvp_1;(u^2-v^2)p_1$ }. Similarly, we can prove that { $af+be;ae+bf;ae-bf;|af-be|$ } = { $(x^2-y^2)p_2;2xyp_2;2uvp_1;(u^2-v^2)p_1$ } = { $ad+bc;ac+bd;|ad-bc|;ac-bd$ }. Let $S=$ { $ad+bc;ac+bd;|ad-bc|;ac-bd$ } and $T=$ { $af+be;ae+bf;ae-bf;|af-be|$ }. Since $S \equiv T$ , $ac+bd$ is the largest element in $S$ and $ae+bf$ is the largest element in $T$ , we must have $ac+bd=ae+bf \Rightarrow (f-d)$ is divisible by $a$ , a contradiction since $0<f-d<a$ . So $k \geq 3$ , the smallest number $M$ therefore would be $5.13.17=1105=33^2+4^2=32^2+9^2=12^2+31^2=23^2+24^2$