15th Swedish Mathematical Society Problems 1975

1.  A is the point (1, 0), L is the line y = kx (where k > 0). For which points P (t, 0) can we find a point Q on L such that AQ and QP are perpendicular?

2.  Is there a positive integer n such that the fractional part of (3 + √5)ⁿ > 0.99?

3.  Show that aⁿ + bⁿ + cⁿ ≥ ab^n-1 + bc^n-1 + ca^n-1 for real a, b, c ≥ 0 and n a positive integer.

4.  P₁, P₂, P₃, Q₁, Q₂, Q₃ are distinct points in the plane. The distances P₁Q₁, P₂Q₂, P₃Q₃ are equal. P₁P₂ and Q₂Q₁ are parallel (not antiparallel), similarly P₁P₃ and Q₃Q₁, and P₂P₃ and Q₃Q₂. Show that P₁Q₁, P₂Q₂ and P₃Q₃ intersect in a point.

5.  Show that n divides 2ⁿ + 1 for infinitely many positive integers n.

6.  f(x) is defined for 0 ≤ x ≤ 1 and has a continuous derivative satisfying |f '(x)| ≤ C |f(x)| for some positive constant C. Show that if f(0) = 0, then f(x) = 0 for the entire interval. 

Solutions

Problem 2

Is there a positive integer n such that the fractional part of (3 + √5)ⁿ > 0.99?

Answer

yes

Solution

Expanding by the binomial theorem we see that (3 + √5)ⁿ + (3 - √5)ⁿ is an integer, but 0 < (3 - √5) < 1, so (3 - √5)ⁿ < 0.01 for sufficiently large n.

Thanks to Suat Namli

Problem 3

Show that aⁿ + bⁿ + cⁿ ≥ ab^n-1 + bc^n-1 + ca^n-1 for real a, b, c ≥ 0 and n a positive integer.

Solution

By AM/GM (aⁿ + bⁿ + bⁿ + ... + bⁿ)/n ≥ ab^n-1. Similarly (bⁿ + cⁿ + cⁿ + ... + cⁿ)/n ≥ bc^n-1 and (cⁿ + aⁿ + aⁿ + ... + aⁿ)/n ≥ ca^n-1. Adding gives the result.

Thanks to Suat Namli

Problem 5

Show that n divides 2ⁿ + 1 for infinitely many positive integers n.

Solution

For n = 3 we have 2ⁿ + 1 = 9 which is a multiple of 3. Now we have 2^3k + 1 = (2^k + 1)(2^2k - 2^k + 1), so if 2^k + 1 = 0 mod 3, then 2^2k = 1 mod 3, so 2^2k - 2^k + 1 = 1 + 1 + 1 = 0 mod 3. So if 3^h divides 2^k + 1, then 3^h+1 divides 2^3k + 1. Hence (by a trivial induction) for n = 3^k, n divides 2ⁿ + 1.

Thanks to Suat Namli