Question

Show that there is no integer a such that
a^2 − 3a − 19 is divisible by 289

Answer 1

problemsolver where do u get all these problems ?

Answer 2

maybe you could show that:
\[a^2-3a-19 = 289k, k\in\mathbb Z\]
is impossible? a proof by contradiction?

Answer 3

ya contradiction

Answer 4

i dont find prblems they find me

Answer 5

LOL

Answer 6

\[a^2-3a-19 = 289k \Rightarrow a^2-3a-(19+289k) = 0\]
Quadratic formula:
\[\frac{3\pm \sqrt{9+4(19+289k)}}{2}\]
to show this has no integer solutions, i just want to show that the discriminant cant be a perfect square......right?

Answer 7

don no wat u mean sry

Answer 8

i want to show that the inside of that square root cant be a perfect square. i think thats what im trying to say lol

Answer 9

ok

Answer 10

80 +1156k inside root

Answer 11

so:
\[9+4(19+289k) = 9+76+1156k = 85+1156k\]
hmmm.....

Answer 12

ah, a 17 factors out, kinda convenient see that 289 is 17^2 <.<
so now we have:
\[17(5+68k)\]

Answer 13

good morning satellite!

Answer 14

i think the idea is to see what happens when you square numbers of the form
\[17j\]
\[17j+1\]
\[17j+2\] etc. but the only reason i think that is because i vaguely remember this, so perhaps this is cheating.

good morning joemath!

Answer 15

now , if the inside of the square root were a perfect square, then (5+68k) should also have a factor of 17 in it:
\[5+68k \equiv 0 \mod 17\]
but, 68 = 17*4, so we have:
\[5+68k = 5+17*4k \equiv 5 \mod 17\]
There is no way for a factor of 17 to come out of that expression. Therefor the inside of the square root will never be a perfect square, and there are no integer solutions blah blah contradiction blah.