MathLabs

Problem 3

Prove that ∑k=0n(2n+12k+1)23k\sum_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k} is not divisible by 55 for any integer n≥0n\ge0.
Step 3 of 5: Compute the norm
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(1−2)2n+1=α−β2,−1=α2−2β2(1-\sqrt2)^{2n+1}=\alpha-\beta\sqrt2,\quad -1=\alpha^2-2\beta^2
Detailed analysis

The conjugate expansion is (1−2)2n+1=α−β2(1-\sqrt2)^{2n+1}=\alpha-\beta\sqrt2. Multiplying the two conjugates gives ((1+2)(1−2))2n+1=(−1)2n+1=−1((1+\sqrt2)(1-\sqrt2))^{2n+1}=(-1)^{2n+1}=-1, while the right side is α2−2β2\alpha^2-2\beta^2. Hence α2−2β2=−1\alpha^2-2\beta^2=-1 in F5\mathbb F_5.