A simple Factorial Diophantine equation (17x^2=9y!+2003)

preview_player
Показать описание
A simple Factorial Diophantine equation (17x^2=9y!+2003)

Factorial Equation | Factorial Diophantine equation | Algebra | A simple Factorial Diophantine equation
Рекомендации по теме
Комментарии
Автор

Similar logic shows that the solution set's finite for mx^n = py! + q if m doesn't divide q, where m, n, p & q are all integers.

tedbagg
Автор

Great, actually looking at mod 4 for y>= 4 is already enough, since we have the nice fact that a^2 is congruent to 0, 1 mod 4 for all natural a.
Taking mod 4 reduces this to x^2 is congruent to y! +3, but y >= 4, this means y! is 0 mod 4, that means x^2 is 3 mod 4 which clearly means theres no positive integer solutions for x.

The above solution is motivated by the fact perfect squares mod 4 have a nice residue. Also y=3 works, and by testing 4, 5, 6…, motivates mod 4.
Nonetheless, great vid.

beautyofmath
Автор

How do you test for divisibility by 17 without using a calculator? I wouldn't have put it past Gauss, but if you have a rule, I would like to have it explained.

zh