Jump to content

Photo

$n\in\mathbb{Z}^+$?: $(n-1)!+(n+1)^2 = (n^2-41)(n^2+49).$

- - - - -

  • Please log in to reply
No replies to this topic

#1
Baoriven

Baoriven

    Thượng úy

  • Điều hành viên OLYMPIC
  • 1425 posts

Tìm tất cả các số nguyên dương $n$ thoả mãn:

$$(n-1)!+(n+1)^2 = (n^2-41)(n^2+49).$$


$$\mathbf{\text{Every saint has a past, and every sinner has a future}}.$$





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users