MathLabs

Problem 4

Let n≥2n\ge2 be a positive integer with divisors 1=d1<d2<⋯<dk=n1=d_1<d_2<\cdots<d_k=n. Prove that d1d2+d2d3+⋯+dk−1dkd_1d_2+d_2d_3+\cdots+d_{k-1}d_k is always less than n2n^2, and determine when it is a divisor of n2n^2.
Step 1 of 3: Pair complementary divisors
Previous stepNext step
In plain words

Pair complementary divisors

S=∑i=1k−1didi+1=n2∑i=1k−11didi+1S=\sum_{i=1}^{k-1}d_id_{i+1}=n^2\sum_{i=1}^{k-1}\frac1{d_id_{i+1}}
Detailed analysis

Since didk−i+1=nd_i d_{k-i+1}=n, reversing the summands gives the displayed identity.