User:JoshuaZ/Draft

Theorem: If there exists $$ n $$ tau such that $$n \equiv a $$ (mod) $$k$$ then there exist arbiritrarily large $$m$$, $$m$$ tau such that $$n \equiv $$ a mod k.

Proof: Assume as above. For any prime $$p$$ such that $$p \equiv 1 $$ (mod) $$ k $$ and $$p$$ not a divisor of $$n$$  then $$m=p^{p-1}n $$ is tau since $$ \tau (p^{p-1}n) = p\tau(n) |p^{p-1} \tau (n) =m)$$. By Dirichlet's Theorem there are arbitrarily large such $$p$$.

Theorem: If $$n$$ is an odd perfect number satisfying 3 is not a divisor of $$ n $$ then are integers $$x,y$$ such that $$3x^2+y^2=n$$ Proof: Assume as above. By Euler's theorem $$n=q^km^2$$ for some$$q \equiv m \equiv 1 $$ (mod 4) and $$m$$ relatively prime to $$q$$. We must have $$q \equiv 1 $$ (mod 3) since otherwise $$3 | \sigma(q^k)|n $$. But then $$q = 3x'^2 + y'^2$$ for some $$x',y'$$. Then $$x=x'q^{\frac{k-1}{2}}m, y=y'q^{\frac{k-1}{2}}m $$ works.