Question

is there a 3-digit number for which\[\overline{abc}=b^2+4ac\]

Answer 1

what's that symbol?

Answer 2

that means abc is a number

Answer 3

like abc=100*a+10b+c ?

Answer 4

yeah

Answer 5

100*a+10b+c = b^2+4ac <-- this seems like determinant of ax^2+bx-c = 0

Answer 6

\[a,b,c \in \text{{0,1,...,9}}\]

Answer 7

a cannot be 0. Otherwise, it's not a 3-digit number.

Answer 8

yeah a can not be 0

Answer 9

RHS is limited

Answer 10

a, b, c are distinct, right?

Answer 11

they can be same

Answer 12

100*a+10b+ (c - b^2-4ac) = 0
(10 * 2a + b)^2 = b^2 - 4a(c - b^2-4ac)

Answer 13

this is quadratic in a

Answer 14

how u bring a^2 to work

Answer 15

100*a+10b+ (c - b^2-4ac) = 0
x^2a+xb + (c - b^2-4ac) = 0, x=10

Answer 16

perhaps we can use quadratic property and uniqueness to reduce things ... looks like this is too long and not my stuff. I should be doing PDE !!

Answer 17

lol...man a is limited

Answer 18

cannot have double value .. so discriminant should be zero.

Answer 19

juantweaver started good

Answer 20

no number exists...the only possible chance is when a=1 but then there is no integer solution for b,c so both sides equal

Answer 21

rewrite equation as \[100*a+c-4ac=b^2-10b\]
then prove that LHS is always positive and RHS is always negative. impossible

Answer 22

i like that proof ^^

Answer 23

that one is nice too juant

Answer 24

only possibility for a is 1
-> only possibility for b -> 8 and 9
-> no integer value for c