A number theory problem by Djordje Veljkovic

Let's rewrite the above equation as a quadratic in $b$ : $b^2 - 6b + (2014 - 9a^2) = 0$ , which has roots:

$b = \frac{6 \pm \sqrt{36 - 4(1)(2014 - 9a^{2})}}{2} = 3 \pm \sqrt{9a^2 - 2005}.$

We require the discriminant to be a perfect square in order for $b \in \mathbb{Z}$ ,

or $9a^2 - 2005 = k^2 \Rightarrow 9a^2 - k^2 = 2005 \Rightarrow (3a + k)(3a - k) = 2005$

which 2005 has eight integer divisors $\pm1, \pm5, \pm401, \pm2005$ . Taking the following systems of equations:

$3a+k = 2005, 3a-k = 1 \Rightarrow a = \frac{1003}{3};$

$3a+k = -1, 3a-k = -2005 \Rightarrow a = -\frac{1003}{3};$

$3a+k = 401, 3a-k = 5 \Rightarrow a = \frac{203}{3};$

$3a+k = -5, 3a-k = -401 \Rightarrow a = -\frac{203}{3}.$

which shows that $a \not\in \mathbb{Z}$ and therefore no integral pair $(a,b)$ exists.

The number $2014$ gives a remainder of $1$ when divided by $3$ . Both $9a^2$ and $6b$ are divisible by $3$ , meaning that $b^2$ must give the remainder of $2$ when divided by $3$ . Any square of an integer is either divisible by $3$ or gives a remainder of $1$ when divided by $3$ , meaning that there are no numbers $a$ and $b$ that fulfill the above condition.