Question

fun question :D

Answer 1

How many integers 'x' are there such that \[x^2 - 3x -19 \] is divisible by 289? :)

Answer 2

28

Answer 3

nd hw did u get that :)

Answer 4

thats wrng btw

Answer 5

ok

Answer 6

no thats not correct :)

Answer 7

don't guess :)

Answer 8

so do u knoe correct answer huh ? ok give me options if u can

Answer 9

yes i knw the answer :D nd this is a subjective question

Answer 10

can u give me options ?

Answer 11

ok
your options are -
a)5
b)9
c)289
d)none of these

Answer 12

u are asking about numbers that are divisible by 289 or only a certain number for x?

Answer 13

m asking that for how many values of x the expression \[x^2 -3x -19 \] is divisible by 289

Answer 14

no one

Answer 15

:) can u give any reason to support your answer?

Answer 16

yes because only 289 is divisible by 289 not any other number ...u can't get 289 by putting any value of x

Answer 17

ganes is typing a reply ...:)

Answer 18

Suppose \( x^2-3x-19\equiv 0\pmod{289}~~\color{red}{\star}\)

Since \(289=17^2\),
it must necessarily be the case that \(x^2-3x-19 \equiv 0\pmod{17}\)

multiply \(4\) through out, complete the square and simplifying gives :
\[(2x-3)^2\equiv 0\pmod{17}\]
so the solution is \(x\equiv 10\pmod{17}\)

Answer 19

plugging that solution in \(\color{red}{\star}\) :
\(x^2-3x-19\equiv (10+17k)^2-3(10+17k)-19 \\
\equiv 100+340k+289k^2-30-51k-19\\
\equiv 51+289k+289k^2\\
\equiv 51\\
\not\equiv 0\pmod{289}
\)

contradiction \(\blacksquare\)

Answer 20

yes :D

Answer 21

im pretty sure there are other simpler ways to solve this
but what i showed you uses the standard method to solve any quadratic congruence... no tricks!

Answer 22

yes :) there is another simple method to solve it.

Answer 23

il let others try :)

Answer 24

ok :)

Answer 25

\[x^2 - 3x - 19 = 289t\]\[\Rightarrow x^2 - 3x - 19 - 289t = 0\]This equation has integral roots. So for that to happen, \(D\) is a perfect square.\[\Rightarrow 9 + 4(19 - 289t) = 85 - 4\cdot 289 t = 17 (5 - 68t)\]is a perfect square. The second factor must be divisible by 17 to make it a perfect square. (And that is only one condition.) But that's not possible, right?

Answer 26

That's pretty clever!

Answer 27

In general, for integral roots of the equation \(x^2 + bx + c = 0\), the discriminant must be a perfect square. Proof:\[ x= \frac{-b\pm \sqrt{b^2 - 4c}}{2}\]Now if \(b\) is even then \(D\) is also even, hence its root which is an integer is also even. So we've shown that the roots will be even. Similarly, we can think about the odd-case.

Answer 28

@Astrophysics I love discriminants.

Answer 29

I like your method more! the discriminant must be a perfect square for the roots to be rational, otherwise we get a nonsimplifying radical in the quadratic formula...

Answer 30

No, it actually means that the roots are not only rational, but integral. :P

Answer 31

clever and neat @ParthKohli

Answer 32

how do you know the top is always even ?

Answer 33

I wrote the proof above. Take it casewise. First, take b to be even, then odd. If we know that D is a perfect square, then the top will be even. Surprising.

Answer 34

Ahh okay, the only rational roots of \(x^2+bx+c\) are integers by rational root theorem

Answer 35

Oh, that's a good insight. Thanks. Why in the world didn't I think about that... >_<

Answer 36

my method -
\[x^2 -3x -19\]\[(x-10)(x+7)+51\]
suppose 289 divides \[x^2 - 3x -19\] so for some integer x. then 17 divides it ...we knw that 17 divides 51 nd also 17 must divide (x-10)(x+7).
Since 17 is prime it must divide (x-10) or (x+7).
nd we see that (x+7-(x-10) = 17 so when 17 divides one of (x-10) or (x+7) it must also divide the other :) so this means 17^2=289 can divide (x-10)(x+7)

it follows that 289 can divide 51 :D which is impossible .

So there is no integer x for which 289 divides \[x^2 - 3x -19\]

Answer 37

Actually i was considering the case \(ax^2+bx+c\) when \(a\ne 1\) somehow and thought ur earlier statement must be wrong

Answer 38

Wow that is a cool one too! now we have 3 different methods

Answer 39

:)

Answer 40

Hmm, it seems like you guys got the same contradiction in the end. Cool answers.

Answer 41

yes first and third methods look more or less similar, but the 3rd solution looks better because it avoids all that congruence mess

Answer 42

2nd solution is my fav so far though

Answer 43

Thanks, @geniusie8.