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Consider the sequence: $$ a_n=x^n+y^n $$ We can write this as a recurrence relation: $$ \begin{align} a_n&=(x+y)\ a_{n-1}-xy\ a_{n-2} \\ a_n&=(x+y)\left(x^{n-1}+y^{n-1}\right)-xy\left(x^{n-2}+y^{n- …

This answer is an improvement on Milo Brandt's answer, and is based on "Formes linéaires en deux logarithmes et déterminants d′interpolation" (M. Laurent, M. Mignotte, and Y. Nesterenko) (which I got …

Essentially what we need to do is find a lower bound on the sum: $$ \sum_{i=1}^{n}\sum_{j=1}^{n}\operatorname{gcd}(i,j) \tag{1} $$ This is the total of the entries in the $n\times n$ table of GCD's. …

The Chinese remainder theorem renders this type of problem easy. However, in this case the answer is even easier: The Chinese remainder theorem tells us that there are an infinite number of number …

The right side increases with $n$, and we know that Therefore, for $n\geq 3$, Now we only have to check $n=2$ and $n=1$ by hand.

(Please tell me why this works, I have no idea!) Hint 2b suggests that there is a relationship between binary representations of coin weights and continued fractions. I assumed that this relationshi …

By brute-force search, yes. I started by searching over all tuples $(r,s,t)$ less than $1000$, stopping at the first example I found: $$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$ …

Answer to part c): Found with the following Python code: from heapq import heappush, heappop best_n = 0 candidates = [(1, [1], set((1,)))] while candidates: prev_max, prev_list, prev_set = heap …

Adding Translating the exponents from decimal to ASCII gives: The title "Goldbach's safe" and name "C. Goldbach" likely refers to: