Deep number theory

${ a }^{ 2 }-440={ b }^{ 2 }\\ \therefore { a }^{ 2 }-{ b }^{ 2 }=440\\ \therefore (a+b)(a-b)=440\\ \\ Let\quad { f }_{ 1 }=(a+b)\quad and\quad { f }_{ 2 }=(a-b)\\ \boxed { \therefore { f }_{ 1 }>{ f }_{ 2 } } ,\boxed { { f }_{ 1 }{ f }_{ 2 }=440 } \\ \\ { f }_{ 1 }+{ f }_{ 2 }=a+b+a-b=2a\\ \boxed { \therefore { f }_{ 1 }+{ f }_{ 2 }\quad is\quad even } \\ \\ 440=2.2.2.5.11={ 2 }^{ 3 }.5.11\\ \\ No.\quad of\quad factors\quad of\quad { 2 }^{ \alpha }.{ 3 }^{ \beta }.{ 5 }^{ \gamma }.{ 7 }^{ \delta }.....\quad is\\ (\alpha +1)(\beta +1)(\gamma +1)(\delta +1).....\\ \therefore No.\quad of\quad factors\quad of\quad { 2 }^{ 3 }.5.11\quad is\\ 4.2.2=16\\ \\ \therefore 440\quad can\quad be\quad written\quad in\quad 16\quad ways\quad i.e.\\ (This\quad is\quad done\quad to\quad satisfy\quad \boxed { { f }_{ 1 }{ f }_{ 2 }=440 } )\\ 1*440\qquad 440*1\\ 2*220\qquad 220*2\\ 4*110\qquad 110*4\\ 5*88\qquad \quad 88*5\\ 8*55\qquad \quad 55*8\\ 10*44\qquad 44*10\\ 11*40\qquad 40*11\\ 20*22\qquad 22*20\\ \\ \boxed { { \because f }_{ 1 }>{ f }_{ 2 } } ,\quad 8\quad ways\quad are\quad eliminated\\ Now\quad we\quad have\quad 8\quad ways\quad left\quad i.e.\\ 440*1\qquad 220*2\qquad 110*4\\ 88*5\qquad \quad 55*8\quad \qquad 44*10\\ 40*11\qquad 22*20\\ \\ \boxed { \because { f }_{ 1 }+{ f }_{ 2 }\quad is\quad even } ,\quad 4\quad ways\quad are\quad eliminated\\ Now\quad we\quad have\quad 4\quad ways\quad left,\quad i.e.\\ Case\quad 1:\quad 220*2\\ Case\quad 2:\quad 110*4\\ Case\quad 3:\quad 44*10\\ Case\quad 4:\quad 22*20\\ \\ So,\quad here\quad we\quad can\quad say\quad that\quad answer\quad is\quad 4.\\ But\quad further\quad explanation....\\ \\ Case\quad 1:\\ a+b=220\\ a-b=2\\ \therefore a+b+a-b=222\\ \therefore 2a=222\\ \therefore a=\boxed { 111 } \\ \\ Case\quad 2:\\ a+b=110\\ a-b=4\\ \therefore a+b+a-b=114\\ \therefore 2a=114\\ \therefore a=\boxed { 57 } \\ \\ Case\quad 3:\\ a+b=44\\ a-b=10\\ \therefore a+b+a-b=54\\ \therefore 2a=54\\ \therefore a=\boxed { 27 } \\ \\ Case\quad 4:\\ a+b=22\\ a-b=20\\ \therefore a+b+a-b=42\\ \therefore 2a=42\\ \therefore a=\boxed { 21 } \\ \\ So\quad we\quad have\quad \boxed { four } \quad values\quad of\quad a\quad which\\ are\quad \boxed { 111 } ,\boxed { 57 } ,\boxed { 27 } \quad and\quad \boxed { 21 } \\ \\ \\ \\ \\ \\ \\ \\ \quad \quad \quad \quad \quad \\$