First notice that $a^2=9555^2+c^2$ can be rewritten in the following form:
$a^2-c^2=9555^2$
$(a+c)(a-c)=9555^2$
Since $9555=3\cdot5\cdot7\cdot7\cdot13$, if we are to arrange this number into the product of two square numbers, we see that we can have a total of $3+3+1+3+1=11$ ways to do so.
Those 11 ways can be formulated in the following manner:
[TABLE="class: grid, width: 800"]
[TR]
[TD]1.
$(a+c)(a-c)=(3\cdot5)^2(7^2\cdot13)^2=15^2\cdot637^2=202997^2-202772^2$[/TD]
[TD]$(a,c)=(202997, 202772)$[/TD]
[/TR]
[TR]
[TD]2.
$(a+c)(a-c)=(3\cdot7)^2(5\cdot7\cdot13)^2=21^2\cdot455^2=103733^2-103292^2$[/TD]
[TD]$(a,c)=(103733, 103292)$[/TD]
[/TR]
[TR]
[TD]3.
$(a+c)(a-c)=(3\cdot13)^2(5\cdot7^2)^2=39^2\cdot245^2=30773^2-29252^2$[/TD]
[TD]$(a,c)=(30773, 29252)$[/TD]
[/TR]
[TR]
[TD]4.
$(a+c)(a-c)=(5\cdot7)^2(3\cdot7\cdot13)^2=35^2\cdot273^2=37877^2-36652^2$[/TD]
[TD]$(a,c)=(37877, 36652)$[/TD]
[/TR]
[TR]
[TD]5.
$(a+c)(a-c)=(5\cdot13)^2(3\cdot7^2)^2=65^2\cdot147^2=12917^2-8692^2$[/TD]
[TD]$(a,c)=(12917, 8692)$[/TD]
[/TR]
[TR]
[TD]6.
$(a+c)(a-c)=(7\cdot13)^2(3\cdot5\cdot7)^2=91^2\cdot105^2=9653^2-1372^2$[/TD]
[TD]$(a,c)=(9653, 1372)$[/TD]
[/TR]
[TR]
[TD]7.
$(a+c)(a-c)=(3\cdot5\cdot13)^2(7^2)^2=195^2\cdot49^2=20213^2-17812^2$[/TD]
[TD]$(a,c)=(20213, 17812)$[/TD]
[/TR]
[TR]
[TD]8.
$(a+c)(a-c)=(3\cdot5\cdot7\cdot7)^2(13)^2=735^2\cdot13^2=270197^2-270028^2$[/TD]
[TD]$(a,c)=(270197, 270028)$[/TD]
[/TR]
[TR]
[TD]9.
$(a+c)(a-c)=(3\cdot5\cdot7\cdot13)^2(7)^2=1365^2\cdot7^2=931637^2-931588^2$[/TD]
[TD]$(a,c)=(931637, 931588)$[/TD]
[/TR]
[TR]
[TD]10.
$(a+c)(a-c)=(3\cdot7\cdot7\cdot13)^2(5)^2=1911^2\cdot5^2=1825973^2-1825948^2$[/TD]
[TD]$(a,c)=(1825973, 1825948)$[/TD]
[/TR]
[TR]
[TD]11.
$(a+c)(a-c)=(5\cdot7\cdot7\cdot13)^2(3)^2=3185^2\cdot3^2=5072117^2-5072108^2$[/TD]
[TD]$(a,c)=(5072117, 5072108)$[/TD]
[/TR]
[/TABLE]Edit:
kaliprasad is so kind to inform me through PM that I missed to consider the number $1$ and hence missed out the case where
$(a+c)(a-c)=(3\cdot5\cdot7\cdot7\cdot13)^2(1)^2=9555^2\cdot1^2=45649013^2-45649012^2$ and the last solution set to the problem would be $(a,c)=(45649013, 45649012)$!
Thank you
kaliprasad so much for your kind gesture!
