Question 252017
Q.3:-Show that there is no integer a such that a^2-3a-19 is divisible by 289? 
<pre><font size = 4 color = "indigo"><b>
Let {{{(a^2-3a-19)/289=k}}}

We need to show that k is not an integer.

Assume, for contradiction, that it is an integer.

{{{a^2-3a-19=289k}}}

{{{a^2-3a-19=289k}}}

{{{a^2-3a-19-289k=0}}}

{{{a^2-3a+(-19-289k)=0}}}

For {{{Ax^2+Bx+C=0}}} to have a rational solution,
the discriminant {{{B^2-4AC}}} must be a perfect 
square.

Here {{{A=1}}}, {{{B=-3}}},  {{{C=-19-288k}}}

discriminant = {{{ (-3)^2-4(-19-289k)=9+76+1156k=85+1156k=17(5+68k)}}}

In order for this discriminant, {{{17(5+68k)}}}, to be a 
perfect square, since it has one factor of 17, the other
factor {{{5+68k}}} must also have a factor of 17.  So there
must be an integer n such that

{{{5+68k=17n}}}

{{{5=17n-68k}}}

{{{5=17(n-4k)}}}

{{{5/17=n-4k}}}

But the right side is an integer but the left side is not.

This contradicts our assumption that the discriminant is a perfect
square.   

Therefore "a" cannot be a rational number.  And since "a" is
not a rational number, it certainly cannot be an integer.

Edwin</pre>