MHB Proving $a^2-3a-19$ is Not Divisible by 289

[kaliprasad]

Show that there is no integer a for which $a^2-3a -19$ is divisible by 289

 

[Opalg]

Show that there is no integer a for which $a^2-3a -19$ is divisible by 289

[sp]Proof by contradiction: Suppose that $a^2-3a -19$ is divisible by $289 = 17^2$. Since $a^2-3a -19 = (a+7)^2 -17(a+4)$, it follows that $(a+7)^2$ must be divisible by $17.$ But $17$ is prime, and therefore $a+7$ must be divisible by $17.$ So $a+7 = 17k$ for some integer $k$. But then $$a^2-3a -19 = (17k-7)^2 - 3(17k-7) - 19 = 289(k^2-k) + 51,$$ which is clearly not a multiple of $289$.[/sp]

 

[Greg]

Show that there is no integer a for which $a^2-3a -19$ is divisible by 289

Spoiler

$$a^2-3a-19=289k$$

$$\Delta=85+1156k=17(5+68k)$$

For $\Delta$ to be a perfect square $$5+68k$$ must have an odd power of $17$ in its prime factorization and so $5+68k$ must be divisible by $17$, but we see

$$5+68k\equiv5\pmod{17}$$

and so $$a^2-3a-19$$ is not divisible by $289$.

 

[kaliprasad]

2 good ans
here is mine

Spoiler

As a first step as we see that $289 = 17^2$

Now $a^2-3a-19 = (a-10)(a+7) + 51$

The 2nd term that is 51 is divisible by 17 and for the 1st term that is product
to be divisible by 17 either (a-10) or (a+7) is divisible by 17. but if one of them is
divisible by 17 then the 2nd one is also divisible by 17.

So 1st term is (a-10)(a+7) is divisible by 289 and 2nd term 51 is not divisible by 289
so sum is not divisible by 289. Or the 2nd term is divisible by 17 and 1st term is not
divisible by 17 so sum is not divisible by 17.

 

[Deveno]

My solution (which is much like Opalg's):

Spoiler

Reducing mod $17$, and using the quadratic formula (since $2 \not\mid 17$):

$a^2 - 3a - 2 = 0$ (mod $17$), therefore:

$a = 2^{-1}[-(-3) \pm \sqrt{(-3)^2 - 4(-2)}]= 9(3) = 10$ (mod $17$).

Using $a = 17k + 10$ in the original equation, we obtain:

$(17k + 10)^2 - 3(17k + 10) - 19 = 289m$

$289k^2 + 289k + 51 = 289m \implies 17(m - k - k^2) = 3$

but $17$ does not divide $3$.