The intended cute solution

The question I meant to ask on Sunday was whether $\sum_{m,n=1}^\infty1/(m^2+n^2)^2$ converges or diverges, so I’ll give my solution to that problem now. In fact, the methodology closely resembles the divergence proof I gave for the alternate sum, even though this sum converges.

Since all terms in the sum are positive, the sum either converges absolutely or diverges (the sum cannot be conditionally convergent). In either case the terms of the sum may be rearranged without affecting the convergence. For suppose the sum diverges but converges for some rearrangement: since the terms of the rearranged sum are still positive, the rearranged sum would have to converge absolutely, and therefore converge for all rearrangements, contradicting the supposed diverging arrangement.

Therefore we can rearrange the terms as we please; in particular, we can sort the terms in decreasing order, which has the effect of grouping together all terms of the form $1/k^2$ where $k$ is a sum of two squares. The term $1/k^2$ will appear in the sum once for each solution of $m^2+n^2=k$ in positive integers $m$, $n$.1

For our purposes it is sufficient to note there are at most $\sqrt{k}$ solutions to $m^2+n^2=k$. This follows since we know that $1\leq m\leq\sqrt{k}$ and for each value of $m$ there is at most one solution; the only possible value of $n$ which could work is $\sqrt{k-m^2}$.

The argument proceeds as follows:

\[ \newcommand{\N}{\mathbb{N}}\begin{align*}
\sum_{m,n\in\N}\frac{1}{(m^2+n^2)^2} &= \sum_{k=2}^\infty\sum_{\substack{m,n\in\N\\m^2+n^2=k}}\frac{1}{k^2} \\
&\leq \sum_{k=2}^\infty\frac{\sqrt{k}}{k^2} \\
&= \sum_{k=2}^\infty\frac{1}{k^{3/2}} \\
&= \zeta(3/2)-1
\end{align*} \]

The final sum converges since it is a $p$-series (truncated). By the comparison test, the sum in question also converges.

  1. The exact number of solutions to $m^2+n^2=k$ is essentially the sum of squares function $r_2(k)$, although since we are exclusively working with solutions in positive numbers the actual number of solutions will be $r_2(k)/4$ if $k$ is not a perfect square and $(r_2(k)-4)/4$ if $k$ is a perfect square.

    Via MathWorld, we see that if $r_2(k)$ is nonzero then it is equal to $4B(k)$, where $B(k)$ is the number of divisors of $k$ solely comprised of primes congruent to $1$ mod $4$. In any case, this shows that the maximum number of times $1/k^2$ can appear in the summation is $d(k)$, the number of divisors of $k$. In Apostol (page 296) it is shown that $d(k)=o(k^\epsilon)$ for any $\epsilon>0$, so this is a fairly slowly growing function.

Another cute solution

Previously I asked if the summation $\sum_{m,n=1}^\infty1/(m^2+n^2)$ converges or diverges. Actually, the I intended the denominator of the summation term to be $(m^2+n^2)^2$. But never mind, let’s solve the problem as given. To do this, we’ll employ the comparison test.

First, note that

\[ m^2+n^2 \leq m^2+2mn+n^2 = (m+n)^2 , \]

so $1/(m^2+n^2)\geq1/(m+n)^2$.

Next, assume the sum converges; since all terms are positive the sum absolutely converges and the terms may be rearranged without affecting its value. In particular, we rearrange the terms in decreasing order, by grouping all terms equal to $1/k^2$ together for each possible value of $k$.

Finally, we use the fact that there are exactly $k-1$ solutions to $m+n=k$ in positive integers $m$, $n$. Putting it all together, the argument goes as follows:

\sum_{m,n\in\N}\frac{1}{m^2+n^2} &\geq \sum_{m,n\in\N}\frac{1}{(m+n)^2} \\
&= \sum_{k=2}^\infty\sum_{\substack{m,n\in\N\\m+n=k}}\frac{1}{k^2} \\
&= \sum_{k=2}^\infty\frac{k-1}{k^2} \\
&\geq \frac{1}{2}\sum_{k=2}^\infty\frac{1}{k} \\
&= \infty

The final inequality just uses $k-1\geq k/2$ for $k\geq2$ and we are left with a harmonic series, which diverges. By the comparison test the original sum diverges; this contradicts the assumption that it converges, so the sum really does diverge, as required.

Next up, I’ll show the question I intended to ask: does $\sum_{m,n=1}^\infty1/(m^2+n^2)^2$ converge or diverge?

Polynomial problem answer

A month ago I posed the problem of showing that a monic $\newcommand{\Z}{\mathbb{Z}}f\in\Z[x]$ is squarefree over $\newcommand{\C}{\mathbb{C}}\C$ if and only if it is squarefree over $\Z$.

One direction is straightforward, although the easiest way to see it is to consider the contrapositive. If $f$ is not squarefree over $\Z$ then its factorization over $\Z$ is of the form $hg^2$ where $h$, $g\in\Z[x]$ and $g$ is nonconstant. Since $f$ can only factor farther in $\C$ it follows that $f$ is also not squarefree over $\C$; in particular any root $\alpha\in\C$ of $g$ will have multiplicity at least $2$ in the factorization of $f$ in $\C$.

Thus if $f$ is squarefree over $\C$ it is also squarefree over $\Z$. Conversely, if $f$ is not squarefree over $\C$ then it has some multiple root $\alpha\in\C$ and its factorization is of the form $k\cdot(x-\alpha)^2$, so $f’=k'(x-\alpha)^2+2(x-\alpha)k$ and $\alpha$ is also a root of $f’$.

Since $f$ and $f’$ are polynomials with integer coefficients with $\alpha$ as a root, the minimal polynomial $g$ of $\alpha$ over $\newcommand{\Q}{\mathbb{Q}}\Q$ divides both $f$ and $f’$. In particular, there must be some $h\in\Q[x]$ such that $f=gh$. Then $f’=g’h+h’g$ and since $g\mid f’$ and $g\mid h’g$ we must have $g\mid g’h$. However, $g\nmid g’$ since $g’$ has a smaller degree than $g$ and is $g’$ is nonzero (as the characteristic of $\Q$ is $0$).

Being a minimal polynomial, $g$ is irreducible, and it is also prime as $\Q[x]$ is a UFD. Thus from $g\mid g’h$ and $g\nmid g’$ we must have $g\mid h$, i.e., $g^2\mid f$ and $f$ is not squarefree over $\Q$. By Gauss’ Lemma the factorization $f=g^2(h/g)$ over $\Q$ may actually be taken to be over $\Z$ (by replacing $g$ and $h$ with their primitive parts), so $f$ is not squarefree over $\Z$, as required.

Note that the requirement that $f$ be monic is necessary (at least, the content of $f$ should be squarefree). For example, if $f:=4$ then $f=2\cdot2$ is not squarefree over $\Z$, but is technically squarefree over $\C$, since $4$ is a unit in $\C$!