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I am doing some exercises on number theory on multiplicative number theoretic functions and I have some problems with the multiplication on sums like the sum $\sum_{d\mid n}(\tau(d))^2$ where $d$ is a divisor of $n$ and the sums like:

$$\sum_{d\mid n}τ(d)^3$$

same thing except that the function $\tau$ is on $3$rd power..I have to prove that:

$$(\sum_{d\mid n}\tau(d))^2=\sum_{d \mid n}\tau(d)^3$$

where the $\tau(n)$ is the number of positive divisors of $n$ (and where the sums range over all positive divisors $d$ of $n$).

darij grinberg
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Jam
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1 Answers1

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Hint: Your two functions are multiplicative. So to prove they are the same, it is enough to show that they are the same for $n$ a prime power $p^k$.

If $n=p^k$, then $$\sum_{d|n}(\tau(d))^3=1^3+2^3+\cdots +(k+1)^3,$$ and $$\sum_{d|n}\tau(d)=1+2+\cdots+(k+1).$$

The fact that $$1^3+2^3+\cdots+m^3=\left(\frac{m(m+1)}{2}\right)^2$$ is a fairly standard result that you may not need to prove. If necessary, a proof can be given by induction.

André Nicolas
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  • References for the fact you mentioned: http://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct and http://math.stackexchange.com/questions/61482/proving-the-identity-sum-limits-k-1n-k3-large-sum-limits-k-1n-k – Martin Sleziak Jun 18 '14 at 12:54