13th Swedish Mathematical Society Problems 1973

1.  log₈2 = 0.2525 in base 8 (to 4 places of decimals). Find log₈4 in base 8 (to 4 places of decimals).

2.  The Fibonacci sequence f₁, f₂, f₃, ... is defined by f₁ = f₂ = 1, f_n+2 = f_n+1 + f_n. Find all n such that f_n = n².

3.  ABC is a triangle with ∠A = 90^o, ∠B = 60^o. The points A₁, B₁, C₁ on BC, CA, AB respectively are such that A₁B₁C₁ is equilateral and the perpendiculars (to BC at A₁, to CA at B₁ and to AB at C₁) meet at a point P inside the triangle. Find the ratios PA₁:PB₁:PC₁.

4.  p is a prime. Find all relatively prime positive integers m, n such that m/n + 1/p² = (m + p)/(n + p).

5.  f(x) is a polynomial of degree 2n. Show that all polynomials p(x), q(x) of degree at most n such that f(x)q(x) - p(x) has the form ∑_2n<k≤3n (a^k + x^k), have the same p(x)/q(x).

6.  f(x) is a real valued function defined for x ≥ 0 such that f(0) = 0, f(x+1) = f(x) + √x for all x, and f(x) < ½ f(x - ½) + ½ f(x + ½) for all x ≥ ½. Show that f(½) is uniquely determined. 

Solutions

Problem 1

log₈2 = 0.2525 in base 8 (to 4 places of decimals). Find log₈4 in base 8 (to 4 places of decimals).

Answer

0.5253

Solution

log₈4 = 1 - log₈2 

Problem 2

The Fibonacci sequence f₁, f₂, f₃, ... is defined by f₁ = f₂ = 1, f_n+2 = f_n+1 + f_n. Find all n such that f_n = n².

Answer

12

Solution

The first few values are f₁ = 1, f₂ = 1, f₃ = 2, f₄ = 3, f₅ = 5, f₆ = 8, f₇ = 13, f₈ = 21, f₉ = 34, f₁₀ = 55, f₁₁ = 89, f₁₂ = 144, f₁₃ = 233, f₁₄ = 377. Note that the only solution to f_n = n² for n ≤ 12 is n = 12. We claim that f_n > n² for n ≥ 13. It is true for n = 13, 14. Suppose it is true for n, n+1, then f_n+2 > (n+1)² + n² = 2n² + 2n + 1. But n²-2n-3 = (n-3)(n+1) > 0, so 2n²+2n+1 > n²+4n+4 = (n+2)², so the result is true for n+2 and hence for all n ≥ 13.

Thanks to Suat Namli

Problem 4

p is a prime. Find all relatively prime positive integers m, n such that m/n + 1/p² = (m + p)/(n + p).

Answer

for p = 2, (m,n) = (1,2) or (1,4) for p > 2, (m,n) = (p²-p-1,p²).

Solution

Multiplying across we get (n-m)p³ = n(n+p). But n is prime to n-m, so n must divide p³. If n = 1, then p³ divides p+1. Contradiction. If n = p, then p² divides 2p, so p = 2, so m = 1. If n = p², then n-m = p+1, so m = p²-p-1. If n = p³, then p³-m = p³+p, which is impossible since m and p are both positive.