Is there a deficient $N$ such that $\sqrt{D(N)} \mid A(N)$ holds, where $D(N)$ is the deficiency and $A(N)$ is the sum of the aliquot parts of $N$? Do there exist deficient numbers $N$ such that $$\sqrt{D(N)} \mid A(N...--prophetes.ai

Is there a deficient $N$ such that $\sqrt{D(N)} \mid A(N)$ holds, where $D(N)$ is the deficiency and $A(N)$ is the sum of the aliquot parts of $N$? Do there exist deficient numbers $N$ such that $$\sqrt{D(N)} \mid A(N)$$ holds, where $$D(N) = 2N - \sigma(N)$$ is the deficiency and $$A(N) = \sigma(N) - N$$ is the sum of aliquot parts of $N$?

Yes, there are infinitely many such numbers $N$.

Take $N=2^k$. Then, we get $$\begin{align}\sigma(N)&=\sigma(2^k)=2^{k+1}-1\lt 2^{k+1}=2N \\\\\\\D(N)&=2^{k+1}-\sigma(2^k)=2^{k+1}-(2^{k+1}-1)=1\end{align}$$ from which $$\sqrt{D(N)} \mid A(N)$$ follows.

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Another example is $N=14$ which is not of the form $2^k$.

We get $$\begin{align}\sigma(14)&=(1+2)(1+7)=24\lt 2\times 14 \\\\\\\D(N)&=D(14)=28-\sigma(14)=28-24=4 \\\\\\\A(N)&=\sigma(14)-14=24-14=10\end{align}$$ from which $$\sqrt{D(N)} \mid A(N)$$ follows.