Interesting Number 294

We need $a^2-b^2=(a+b)(a-b)=123456789$

From this, we can say that $a>b$ .

Thus, the solutions would be of the type $((y+x)/2,(y-x)/2)$ , where $xy=123456789$ with $y>x$

Factorizing $123456789$ we find

$123456789 = 3^2\times 3607 \times 3803$

From which we can get six solutions for $(x,y)$ as $(1,123456789),\\ (3,3\times 3607\times 3803), \\ (3^2,3607 \times 3803),\\ (3607, 3^2 \times 3803),\\ (3803, 3^2 \times 3607),\\ (3 \times 3607, 3\times 3803)$

There are six solutions for $(a,b)$

$(61728395,61728394),\\ (20576133,20576130),\\ (6858715,6858706),\\ (18917,15310),\\ (18133,4330),\\ (11115,294)$

123456789 = 3^2 * 3607 * 3803. Thus the total number of factors are 12. Therefore the number of pairs are 12/2 which is 6 .
Is there a way to factorize the number without using wolframalpha? (I had to unfortunately use it)

(a + b)(a - b) = 123456789=3^2 X 3607 X 3803
So, number of divisors of 123456789 = (2 + 1)(1 + 1)(1 + 1) =12.
So, (a + b) can have 12 values. But as a, b are positive, a + b > a - b.
So, we get only 6 values of (a + b) and each value of (a + b) we get a unique value of
(a -b).
So, we get 6 ordered pairs (a, b).