SOLUTION: Q.3:-Show that there is no integer a such that a^2-3a-19 is divisible by 289?

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Question 252017: Q.3:-Show that there is no integer a such that a^2-3a-19 is divisible by 289?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Q.3:-Show that there is no integer a such that a^2-3a-19 is divisible by 289? 

Let %28a%5E2-3a-19%29%2F289=k

We need to show that k is not an integer.

Assume, for contradiction, that it is an integer.

a%5E2-3a-19=289k

a%5E2-3a-19=289k

a%5E2-3a-19-289k=0

a%5E2-3a%2B%28-19-289k%29=0

For Ax%5E2%2BBx%2BC=0 to have a rational solution,
the discriminant B%5E2-4AC must be a perfect 
square.

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

discriminant = +%28-3%29%5E2-4%28-19-289k%29=9%2B76%2B1156k=85%2B1156k=17%285%2B68k%29

In order for this discriminant, 17%285%2B68k%29, to be a 
perfect square, since it has one factor of 17, the other
factor 5%2B68k must also have a factor of 17.  So there
must be an integer n such that

5%2B68k=17n

5=17n-68k

5=17%28n-4k%29

5%2F17=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