14th Swedish Mathematical Society Problems 1974

1.  Let a_n = 2^n-1 for n > 0. Let b_n = ∑_r+s≤n a_ra_s. Find b_n - b_n-1, b_n - 2b_n-1 and b_n.

2.  Show that 1 - 1/k ≤ n(k^1/n - 1) ≤ k - 1 for all positive integers n and positive reals k.

3.  Let a₁ = 1, a₂ = 2^a₁, a₃ = 3^a₂, a₄ = 4^a₃, ... , a₉ = 9^a₈. Find the last two digits of a₉.

4.  Find all polynomials p(x) such that p(x²) = p(x)² for all x. Hence find all polynomials q(x) such that q(x² - 2x) = q(x-2)².

5.  Find the smallest positive real t such that x₁ + x₃ = 2t x₂, x₂ + x₄ = 2t x₃, x₃ + x₅ = 2t x₄ has a solution x₁, x₂, x₃, x₄, x₅ in non-negative reals, not all zero.

6.  For which n can we find positive integers a₁, a₂, ... , a_n such that a₁² + a₂² + ... + a_n² is a square? 

Solutions

Problem 2

Show that 1 - 1/k ≤ n(k^1/n - 1) ≤ k - 1 for all positive integers n and positive reals k.

Solution

AM/GM applied to k, 1, ... , 1 gives k^1/n ≤ (k+n-1)/n, so n(k^1/n - 1) ≤ k-1, which is second inequality.

AM/GM applied to k^(n+1)/n, k^(n+1)/n, ... , k^(n+1)/n, 1 (n+1 terms in all) gives k ≤ (n k^(n+1)/n + 1)/(n+1), or (n+1)k - 1 ≤ n k^(n+1)/n or 1 - 1/k ≤ n k^1/n - n, which is the first inequality.

Thanks to Suat Namli

Problem 3

Let a₁ = 1, a₂ = 2^a₁, a₃ = 3^a₂, a₄ = 4^a₃, ... , a₉ = 9^a₈. Find the last two digits of a₉.

Answer

21

Solution

Working mod 100, we find 9¹ = 9, 9² = 81, 9³ = 29, 9⁴ = 61, 9⁵ = 49, 9⁶ = 41, 9⁷ = 69, 9⁸ = 21, 9⁹ = 89, 9¹⁰ = 1, 9¹¹ = 9, so periodic period 10. Thus we only need to find the last digit of a₈. Working mod 10 we find 8¹ = 8, 8² = 4, 8³ = 2, 8⁴ = 6, 8⁵ = 8, 8⁶ = 4, so periodic period 4. Thus we need to find a₇ mod 4. We find 7¹ = 3, 7² = 1, 7³ = 3, so periodic period 2. Now a6 is obviously even, so a₇ = 1 mod 4. Hence a₈ = 8 mod 10. Hence a₉ = 21 mod 100. 

Problem 5

Find the smallest positive real t such that x₁ + x₃ = 2t x₂, x₂ + x₄ = 2t x₃, x₃ + x₅ = 2t x₄ has a solution x₁, x₂, x₃, x₄, x₅ in non-negative reals, not all zero.

Answer

1/√2

Solution

It is easy to check that x₁ = x₅ = 0, x₂ = x₄ = 1, x₃ = √2, t = 1/√2 works.

Suppose for some t we have a solution x_i. Then 0 = 2tx₂ + 2tx₄ - 4t²x₃ = x₁ + 2x₃ + x₅ - 4t²x₃. Hence x₃ = 0 or 4t² - 2 = (x₁ + x₅)/x₃ ≥ 0. If x₃ = 0, then x₂ + x₄ = 0, so x₂ = x₄ = 0. Hence x₁ + x₃ = 0 and x₃ + x₅ = 0, so all x_i = 0. Contradiction. So we must have 4t² - 2 ≥ 0 and hence t ≥ 1/√2. 

Problem 6

For which n can we find positive integers a₁, a₂, ... , a_n such that a₁² + a₂² + ... + a_n² is a square?

Answer

all positive n

Solution

Take a₁ = 3, a_n+1 = (a₁² + a₂² + ... + a_n² - 1)/2. For put b_n = a₁² + a₂² + ... + a_n². We show by induction that bn is an odd square. True for 1. Suppose true for n, then a_n+1 = (b_n - 1)/2, which is even, since odd squares are 1 mod 4. Hence b_n+1 = b_n + a_n+1² = (b_n² - 2b_n + 1 + 4b_n)/4 = ((b_n + 1)/2)², which is an odd square.

Thanks to Suat Namli