Another Tau question

[Jon Perry]

A000594

Tau(n) is only odd for n=(2k+1)^2.

I spotted this by inspection, and failed to prove it.

However, if true, this tells us that Tau(p^2k)!=0

Furthermore, we could then easily determine the mod4 of Tau(p^2),as:

Tau(p).Tau(p)=Tau(p^2)+p^11.

Jon Perry
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