Irish Intervarsity Mathematics Competition 2006

University College Cork

9.30–12.30 Saturday 4^{th} March 2006

1. Let [x] denote the greatest integer not exceeding the real number x.

Prove that

[\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2}] = [\sqrt{9n + 8}]

for all positive integers n. Answer: Replacing n by n + 1, we have to show that

[\sqrt{n − 1} + \sqrt{n} + \sqrt{n + 1}] = [\sqrt{9n − 1}]

for all integers n ≥ 2. If f(x) = \sqrt{x} then

f′(x) = \frac{1}{2} \frac{1}{\sqrt{n}}, f′′(x) = −\frac{1}{4} \frac{1}{\sqrt{n}^{3}} .

By the Mean Value Theorem

\sqrt{n − 1} − \sqrt{n} = \frac{1}{2} \frac{1}{\sqrt{n − θ}}, \sqrt{n} − \sqrt{n + 1} = \frac{1}{2} \frac{1}{\sqrt{n + ϕ}},

where 0 < θ, ϕ < 1. Hence

\sqrt{n − 1} + \sqrt{n} + \sqrt{n + 1} = 3\sqrt{n} + \frac{1}{2} (\frac{1}{\sqrt{n − θ} − \sqrt{n + ϕ}})

= 3\sqrt{n} − \frac{1}{4} \frac{θ + ϕ}{\sqrt{n + ψ}^{3}} ,

where |ψ| < 1. Similarly

\sqrt{9n − 1} = 3\sqrt{n − 1/9}

= 3\sqrt{n} − \frac{1}{6} \frac{1}{\sqrt{n − χ}},

where 0 < χ < 1/9. Now

\frac{1}{4} \frac{θ + ϕ}{\sqrt{n + ψ}^{3}} < \frac{1}{2} \frac{1}{\sqrt{n − 1}^{3}} ,

while

\frac{1}{6} \frac{1}{\sqrt{n − χ}} > \frac{1}{6} \frac{1}{\sqrt{n}}.

But

\frac{1}{2} \frac{1}{\sqrt{n − 1}^{3}} ≤ \frac{1}{6} \frac{1}{\sqrt{n}}

if

3 ≥ n − 1,

ie

n ≥ 4.

Thus

\sqrt{9n − 1} ≤ \sqrt{n − 1} + \sqrt{n} + \sqrt{n + 1}

if n ≥ 4, Note that

3\sqrt{n} < \sqrt{9n + 8} < 3\sqrt{n + 1}.

For positive x, the function f(x) = \sqrt{x} is convex upwards, since f′′(x) < 0. Hence

\sqrt{n} + \sqrt{n + 2} < 2\sqrt{n + 1}.

Thus

3\sqrt{n} < \sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} < 3\sqrt{n}.

So the two nearest integers must be equal unless

[3\sqrt{n + 1}] > [3\sqrt{n}],

ie

[\sqrt{9n + 9}] > [\sqrt{9n}].

This can only happen if 9n is near to a square. More precisely, if

(m − 1)^{2} ≤ 9n < m^{2}

then

[\sqrt{9n + 9}] = [\sqrt{9n}]

unless

9n = m^{2} − r,

where 0 ≤ r ≤ 9.

2. The sides BC, CA, AB of the triangle ABC have lengths a, b, c, respec-

tively, and satisfy

b^{2} − a^{2} = ac, c^{2} − b^{2} = ab.

Determine the measure of the angles of ABC. Answer: We have

cos A = \frac{b^{2} + c^{2} − a^{2}}{2bc} ,

etc. Let

\frac{b}{a} = t.

Then

t^{2} − 1 = c/a, c^{2}/a^{2} = t^{2} + t.

Thus

(t^{2} − 1)^{2} = t^{2} + t.

Taking out the factor t + 1,

(t − 1)(t^{2} − 1) = t,

ie

t^{3} − t^{2} − 2t + 1 = 0.

Also

\frac{c}{a} = t^{2} − 1, \frac{c^{2}}{a^{2}} = t^{2} + t.

Thus

cos A = \frac{t^{2} + t^{2} + t − 1}{2t^{2}(t + 1)}

= \frac{2t − 1}{2t^{2}}.

Similarly

cos B = \frac{a^{2} + c^{2} − b^{2}}{2ac}

= \frac{1 + t^{2} + t − t^{2}}{2(t^{2} − 1)}

= \frac{t + 1}{2(t^{2} − 1)},

and

cos C = \frac{a^{2} + b^{2} − c^{2}}{2ab}

= \frac{1 + t^{2} − t^{2} − t}{2t}

= \frac{1 − t}{2t}.

3. Prove that the polynomial

x^{100} + 33x^{67} + 67x^{33} + 101

cannot be factored as the product of two polynomials of lower degree with integer coefficients. Answer: Suppose the polynomial does factorise, say

f(x) = u(x)v(x),

where u(x), v(x) are monomial (have leading coefficients 1). Then the constant coefficients in u(x), v(x) have product 101, and so are ±1, ±101. We may suppose that u(x) has constant coefficient ±1. Then the product of the roots of u(x) is ±1, and so some root α has

|α| ≤ 1.

But then

|α^{100} + 33α^{67} + 67α^{33}| ≤ 1 + 33 + 67 = 101,

with equality only if |α| = 1 and

α^{100} = α^{67} = α^{33}.

In fact, if

α^{100} + 33α^{67} + 67α^{33} + 101 = 0

then

α^{100} = α^{67} = α^{33} = −1.

But

α^{33} = −1 ⟹ α^{99} = −1.

and

α^{100} = −1, α^{99} = −1 ⟹ α = 1
⟹ α^{33} = 1,

which is a contradiction.

4. Let S be the set of all complex numbers of the form a + bi, where a and b are integers and i = \sqrt{−1}. Let A be a 2 × 2 matrix with entries in S and suppose A has determinant 1 and that

A^{n} = I.

for some positive integer n, where I is the identity matrix.

Prove that

A^{12} = I.

Answer: The eigenvalues ω_{1}, ω_{2} of A must satisfy

ω^{n} = 1.

Also

ω_{1}ω_{2} = det A = 1.

Thus

ω_{2} = ω_{1}^{−1} = \overline{ω_{1}}.

Hence

tr A = ω_{1} + ω_{2} ∈ ℝ.

Since

tr A = a + d ∈ S,

it follows that

tr A ∈ ℤ.

Moreover,

|tr A| ≤ |ω_{1}| + |ω_{2}| = 2.

Hence

tr A = 0, ±1, ±2.

Note that A is semisimple (diagonalisable) in all cases, since it satisfies the separable equation x^{n} − 1 = 0.

If tr A = 0 then

ω_{2} = −ω_{1} = ω_{1}^{−1} ⟹ ω_{1}, ω_{2} = ±i.

In this case

A^{4} = I.

If tr A = ±2 then

ω_{1} = ω_{2} = ±1,

and

A = ±I,

so that

A^{2} = I.

Finally, if tr A = ±1 then its characteristic equation is

x^{2} ∓ x + 1 = 0.

Thus

ω_{1}, ω_{2} = −ω, −ω^{2} or ω, ω^{2},

where

ω = \frac{1 + \sqrt{−3}}{2} = e^{2π/3},

in which case,

A^{6} = I.

So

A^{12} = I

in all cases.

5. Evaluate

∫_{0}^{π/2} x cot(x) dx.

Answer: Let x = 2t. Then

I = 4 ∫_{0}^{π/4} t cot(2t)dt

= 4 ∫_{0}^{π/4} t \frac{tan^{2} t − 1}{2 tan t} dt

= 2 ∫_{0}^{π/4} t tan t dt − 2 ∫_{0}^{π/4} t cot t dt

= 2 ∫_{0}^{π/4} t tan t dt − 2(I − ∫_{π/4}^{π/2} t cot t dt)

= −2I + 2 ∫_{0}^{π/4} t tan t dt + 2 ∫_{π/4}^{π/2} t cot t dt.

Changing variable t ↦ π/2 − t in the second integral,

3I = 2 ∫_{0}^{π/4} t tan t dt + 2 ∫_{0}^{π/4}(π/2 − t) cot t dt

= π ∫_{0}^{π/4} tan t dt

= π[− log cos t]_{0}^{π/4}

= −π log(1/\sqrt{2})

= \frac{π log 2}{2}.

Hence

I = \frac{π log 2}{6}.

6. Let x and y be positive integers and suppose that

k = \frac{x^{2} + y^{2} + 6}{xy}

is also an integer. Prove that k is the cube of an integer.

Answer: We have

x^{2} + y^{2} + 6 − kxy = 0.

Regarding this as a quadratic in x, from one solution (x, y) we get another integer solution (x_{1}, y) where

xx_{1} = y^{2} + 6,

ie

x_{1} = \frac{y^{2} + 6}{x}.

Similarly regarding the equation as a quadratic in y, from one solution (x, y_{1}) we get another integer solution (x, y_{1}) with

y_{1} = \frac{x^{2} + 6}{y}.

Now let (x, y) be a solution (for a given k) with minimal x, and minimal y for that x. Then

x ≤ y,

and

\frac{x^{2} + 6}{y} ≥ y,

ie

y^{2} ≤ x^{2} + 6.

Thus

x^{2} ≤ y^{2} ≤ x^{2} + 6.

Note that gcd(x, y) = 1. For if p is prime,

p | x, y ⟹ p^{2} | xy ⟹ p^{2} | k.

In particular x ≠ y unless x = 1, in which case

y^{2} + 7 = ky ⟹ y | 7 ⟹ y = 1, k = 8 or y = 7, k = 8.

If y ≠ x then y ≥ x + 1 and so

(x + 1)^{2} ≤ x^{2} + 6 ⟹ 2x ≤ 5 ⟹ x = 1 or 2.

We have already dealt with the case x = 1. It only remains to consider x = 2. In this case

y^{2} ≤ x^{2} + 6 and y > x ⟹ y = 3.

But (x, y) = (2, 3) does not give integral k.

We conclude that there is only a solution when k = 8, which is a cube.