Inspired by Arjen : Difference of square in 3 distinct ways

Since we are looking for $N$ that can be written as difference of two positive integers in 3 different ways so we note that $N$ must have at least 6 different divisors. Let's call them $a,b,c,d,e f$ where $a>b , c> d$ and $e> f$ . Now

$N = \begin{cases} a^2 -b^2 = (a-b)(a+b) =r_1 \times r_2 \\c^2 -d^2=(c-d)(c+d)= r_3 \times r_4 \\ e^2 -f^2=(e-f)(e+f) = r_5\times r_6 \end{cases}$ Further note that the divisors has an integer solution if only if they share the same parity ( even or odd ). Since(for instance) : $a = \frac{1}{2}\,(r_1+r_2) : b = \frac{1}{2} \,(r_2 -r_1)$ .

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case 1: if the all the divisors of $N$ shares the odd parity then $N$ should only have $6$ different divisors and prime factor $2$ shouldn't exist however the prime factors between 2 and 11(inclusive) can exist and we write as : $N = (3)^{q_1} (5)^{q_2}. (7)^{q_3}.\times (11)^{q_4} \\ d(N)\\ = (1+q_1).(1+q_2).\cdots.(1+q_4) =6$ . Now we need focus the total prime factors $(q_1)$ of $3$ in $N$ .

If $q_1 =1, q_2 =2 \implies N = 3^1.5^2 ={\color{#3D99F6}75}$ $q_1 =2 , q_2 = 1\implies N = 3^2 .5 ={\color{#3D99F6} 45}$ . Also $q_1 = 2 \quad \begin{cases} q_4 =1 \implies N = 3^2.11^1={\color{#3D99F6}99} \\ q _ 4=1 \implies N = 3^2.7 = {\color{#3D99F6}63} \end{cases}$

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case2: If all the divisors of $N$ are even numbers sharing the even parity there are more that $6$ divisors and exist only prime factors of $2 ,3,5$ such that $N = 2^{q_1}\times 3^{q_2}\times 5^{q_3}\\ d(N) = 2^{q_1}\times 3^{q_{3}}\times 5_{q_{3}} = 10, 12 \qquad d(N) \neq 8$ If $q_1 = 4 \begin{cases} q_ 2 = 1 \\N = 2^4.3^1 ={\color{#3D99F6} 48} \\ q_3 = 1 \implies N = 2^4.5 \implies N ={\color{#3D99F6} 80}\end{cases}$ For $d(N <100 ) = 12$ will be true if and only if $q_1 = 3$ and $q_3 =2$ hence $N = 2^3.3^2 ={\color{#3D99F6}72}$ as for $q_1 > 4$ $N > 100$ .

Therefore, total required values of $N = 7$ that are $N = 45,48,63,72,75,80,99$ Bonus also can be processed in the same manner.