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## imqwerty one year ago fun question :D

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1. imqwerty

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

2. anonymous

28

3. imqwerty

nd hw did u get that :)

4. imqwerty

thats wrng btw

5. anonymous

ok

6. imqwerty

no thats not correct :)

7. imqwerty

don't guess :)

8. anonymous

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

9. imqwerty

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

10. anonymous

can u give me options ?

11. imqwerty

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

12. anonymous

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

13. imqwerty

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

14. anonymous

no one

15. imqwerty

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

16. anonymous

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

17. anonymous

ganes is typing a reply ...:)

18. ganeshie8

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 $$x\equiv 10\pmod{17}$$

19. ganeshie8

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 $$\blacksquare$$

20. imqwerty

yes :D

21. ganeshie8

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!

22. imqwerty

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

23. ganeshie8

il let others try :)

24. imqwerty

ok :)

25. ParthKohli

$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?

26. ganeshie8

That's pretty clever!

27. ParthKohli

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.

28. ParthKohli

@Astrophysics I love discriminants.

29. ganeshie8

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...

30. ParthKohli

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

31. imqwerty

clever and neat @ParthKohli

32. ganeshie8

how do you know the top is always even ?

33. ParthKohli

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.

34. ganeshie8

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

35. ParthKohli

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

36. imqwerty

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 $x^2 - 3x -19$

37. ganeshie8

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

38. ganeshie8

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

39. imqwerty

:)

40. ParthKohli

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

41. ganeshie8

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

42. ganeshie8

2nd solution is my fav so far though

43. ParthKohli

Thanks, @geniusie8.

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