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 4 of 5: Zero would force a non-residue
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α=0  ⟹  2β2=1  ⟹  β2=3(mod5)\alpha=0\implies 2\beta^2=1\implies\beta^2=3\pmod5
Detailed analysis

If α=0\alpha=0, the norm equation gives 2β2=12\beta^2=1, hence β2=2−1=3\beta^2=2^{-1}=3 modulo 55. But the quadratic residues modulo 55 are 0,1,40,1,4, so 33 is not a square. This contradiction proves α≠0\alpha\ne0.