Inverse trigonometry or number theory

Geometry Level 4

$\large \text{arccot} (8)=\arctan (a)-\arctan(b)$

If the ordered pair of natural numbers $(a_1,b_1)$ and $(a_2,b_2)$ satisfy the equation above, find the minimum value of $a_1+b_1+a_2+b_2$ .

1 solution

Brian Charlesworth
Jun 1, 2015

We have that $\arctan(a) - \arctan(b) = \arctan\left(\dfrac{a - b}{1 + ab}\right) = \text{arccot} \left(\dfrac{1 + ab}{a - b}\right)$

for $ab \gt 1,$ a condition which will be met here since we are looking for natural (and clearly distinct) numbers $a,b.$ Thus we require that

$\dfrac{1 + ab}{a - b} = 8 \Longrightarrow 1 = 8a - 8b - ab \Longrightarrow 65 = (8 + a)(8 - b).$

Now since $65 = 13*5 = 65*1,$ and since $(8 + a) \gt (8 - b) \gt 0,$ we can have either

Thus $a_{1} + b_{1} + a_{2} + b_{2} = 5 + 3 + 57 + 7 = \boxed{72}.$

Yes. SFFT does the trick.

Bonus question : Is is true that there will exist an integer solution for $a,b$ with integer $n$ for the general equation below?

$\large \text{arccot} (n)=\arctan (a)-\arctan(b)$

Re Bonus question: In general, for integer $n \ge 1,$ we can find suitable $a,b$ by solving

$(n + a)(n - b) = n^{2} + 1.$

Depending upon how $n^{2} + 1$ factors we can have multiple solution pairs $(a,b),$ but at the very least we will always have

$n + a = n^{2} + 1, n - b = 1 \Longrightarrow a = n^{2} - n + 1, b = n - 1,$

giving us the identity $\text{arccot}(n) = \arctan(n^{2} - n + 1) - \arctan(n - 1).$

Brian Charlesworth - 6 years ago

Since there are exactly two solutions , then what is the point adding the word 'minimum'? @Calvin Lin @Brian Charlesworth

Ankit Kumar Jain - 3 years, 3 months ago

That was what the creator wanted.

I'm guessing that they didn't want to give away that there are exactly two solutions, so that you work on figuring our what all of them are.

Calvin Lin Staff - 3 years, 3 months ago