4036 Form: Fill & Download for Free

Writing the left side as a single fraction, we get [math]\frac{a+b}{ab}=\frac{3}{2018}[/math]. Since [math]\frac{3}{2018}[/math] is a fraction in its simplest form, any equivalent fraction must be of the form [math]\frac{3k}{2018k}[/math] where [math]k[/math] is an integer, but not necessarily positive. Therefore we have the simultaneous equations [math]a+b=3k[/math] and [math]ab=2018k[/math]. We can rewrite this as [math]a+\frac{2018k}{a}=3k[/math], which boils down to the quadratic equation [math]a^2-3ak+2018k=0[/math]. Using the quadratic formula, we find[math]\displaystyle a = \frac{1}{2} \left(3k \pm \sqrt{9k^2-8072k} \right) \tag*{(1)}[/math]I know you say [math](a,b)[/math] is an ordered pair, but to save time I’ll assume [math]a[/math] is the larger integer and [math]b[/math] is the smaller — at the end, we’ll just remember the pairs can be either way round. This means we pick the [math][/math][math]+[/math] sign for [math]a[/math] and [math]-[/math] sign for [math]b[/math]. Since [math]a \in \mathbb{Z}[/math], the right side of [math](1)[/math] must be an integer. Therefore [math]9k^2-8072k=x^2[/math] must be a perfect square. Multiplying through by [math]9[/math] and rearranging a little shows[math]\displaystyle (9k-4036)^2 - (3x)^2 = 16289296 \tag*{}[/math]Writing [math]16289296[/math] in its prime factorisation form and making the substitutions [math]\alpha = 9k-4036[/math] and [math]\beta = 3x[/math] shows[math]\displaystyle \alpha^2 - \beta^2 = (\alpha+\beta)(\alpha-\beta) = 2^4 \times 1009^2 \tag*{}[/math]Since each bracket is an integer, we can split into a finite number of simultaneous equations, where [math]\alpha+\beta[/math] and [math]\alpha-\beta[/math] are a factor pair of [math]16289296[/math]. To save time, we only need consider the cases when [math]\alpha+\beta[/math] is the bigger factor, and since each bracket has the same parity we can eliminate the cases when one factor is odd. I’ll walk you through one of the equations so you get the idea, but I won’t do them all here. Consider the case when [math]\alpha+\beta=8144648[/math] and [math]\alpha-\beta=2[/math]. Then [math]\alpha=4072325[/math]. Therefore [math]k=\frac{\alpha+4036}{9}=452929[/math]. Plugging this back into [math](1)[/math] shows [math]a=1358114[/math] and [math]b=673[/math]. You can use exactly the same process to solve the other equations (only some of which admit integer solutions), and make sure you remember the cases when [math]\alpha+\beta[/math] and [math]\alpha-\beta[/math] are negative.The only integer pairs are[math]\displaystyle (a,b) = (1358114,673), [/math][math][/math][math]\, (340033,674), [/math][math][/math][math]\, (2018,1009), [/math][math][/math][math]\, (672, -678048) \tag*{}[/math]Finally, don’t forget each pair can be either way round!

Since [math]2017 \equiv 2^2\pmod{11}[/math] and [math]2^5 \equiv -1\pmod{11}[/math], we have[math]2017^{2018} \equiv 2^{4036} = 2 \cdot \big(2^5\big)^{807} \equiv 2 \cdot (-1) \pmod{11}[/math].The remainder is [math]9[/math]. [math]\blacksquare[/math]

We want to evaluate[math]I = \displaystyle \int \dfrac{2017 x^{2016} [/math][math][/math][math]+ 2018 x^{2017}}{1 [/math][math][/math][math]+ x^{4034} [/math][math][/math][math]+ 2 x^{4035} [/math][math][/math][math]+ x^{4036}}\,dx. \tag*{}[/math]To this end, we rewrite it as[math]I = \displaystyle \int \dfrac{2017 x^{2016} [/math][math][/math][math]+ 2018 x^{2017}}{1 [/math][math][/math][math]+ (x^{2017} [/math][math][/math][math]+ x^{2018})^2} \,dx. \tag*{}[/math]Letting [math]w = x^{2017} [/math][math][/math][math]+ x^{2018}[/math], we obtain[math]\begin{align*} I &= \displaystyle \int \dfrac{1}{1 [/math][math][/math][math]+ w^2} \,dw\\ &= \arctan{w} [/math][math][/math][math]+ C\\ &= \arctan(x^{2017} [/math][math][/math][math]+ x^{2018}) [/math][math][/math][math]+ C. \end{align*} \tag*{}[/math]