fun question :D

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fun question :D

Mathematics
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How many integers 'x' are there such that \[x^2 - 3x -19 \] is divisible by 289? :)
28
nd hw did u get that :)

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thats wrng btw
ok
no thats not correct :)
don't guess :)
so do u knoe correct answer huh ? ok give me options if u can
yes i knw the answer :D nd this is a subjective question
can u give me options ?
ok your options are - a)5 b)9 c)289 d)none of these
u are asking about numbers that are divisible by 289 or only a certain number for x?
m asking that for how many values of x the expression \[x^2 -3x -19 \] is divisible by 289
no one
:) can u give any reason to support your answer?
yes because only 289 is divisible by 289 not any other number ...u can't get 289 by putting any value of x
ganes is typing a reply ...:)
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 \(x\equiv 10\pmod{17}\)
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 \(\blacksquare\)
yes :D
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!
yes :) there is another simple method to solve it.
il let others try :)
ok :)
\[x^2 - 3x - 19 = 289t\]\[\Rightarrow x^2 - 3x - 19 - 289t = 0\]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?
That's pretty clever!
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.
@Astrophysics I love discriminants.
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...
No, it actually means that the roots are not only rational, but integral. :P
clever and neat @ParthKohli
how do you know the top is always even ?
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.
Ahh okay, the only rational roots of \(x^2+bx+c\) are integers by rational root theorem
Oh, that's a good insight. Thanks. Why in the world didn't I think about that... >_<
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 \[x^2 - 3x -19\]
Actually i was considering the case \(ax^2+bx+c\) when \(a\ne 1\) somehow and thought ur earlier statement must be wrong
Wow that is a cool one too! now we have 3 different methods
:)
Hmm, it seems like you guys got the same contradiction in the end. Cool answers.
yes first and third methods look more or less similar, but the 3rd solution looks better because it avoids all that congruence mess
2nd solution is my fav so far though
Thanks, @geniusie8.

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