Equivalence of two multisets of natural numbers

Question

I want to show that the two multisets of natural numbers given by :

$\{4m^2+(2n+1)^2\}$ for $m,n \in \mathbb{Z}_{\ge 0}$

and

$\{2(k+l+1/2)^2+2(l+1/2)^2\}$ for $k,l \in \mathbb{Z}_{\ge 0}$

are equal as multisets (i.e. allowing repetitions). These two multisets arise as spectra of some domains that I suspect to be isospectral. One can compute the first few elements, they both start out with

$\{1,5,9,13,17,25,25,29, ... \}$

and numerically they agree further along as well.

My first approach was to assume that $m,n$ can be written as linear combinations of $k,l$ and vice versa but if such a transformation exists, the $2$-by-$2$ matrix would have to have only positive integer entries and its inverse would have to have only positive integer entries as well and the only matrices with these properties are permutation matrices which don't work here.

Answer 1

$\forall k,l \in \mathbb{N}$,

$$ 2(k+l+\frac{1}{2})^2+2(l+\frac{1}{2})^2=2(k^2+l^2+\frac{1}{4}+k+l+2kl)+2(l^2+l+\frac{1}{4})$$

$$ = 2k^2+4l^2+2k+4l+4kl+1$$

$$ = k^2+4l^2+1+4kl+4l+2k+k^2$$

$$ = (k+2l+1)^ 2+k^2$$

When $k=2m$, the expression becomes, $(2(m+l)+1)^2+4m^2$.

Now $l$ can describe $\mathbb{N}$ for $2(m+l)+1$ to describe the set of odd numbers.

When $k=2n+1$, the expression becomes $ (2(n+l+1))^2+(2n+1)^2$.

The remaining $l$ allows all the no null even numbers.

Answer 2

At first notice that: $$ \{ 4m^2+(2n+1)^2 \} _{m,n \in \mathbb{Z}_{\ge 0}} = \{ 4m^2+(2n+1)^2 \} _{m,n \in \mathbb{Z}} ; $$

also note that:

$$ \{ 2(k+l+1/2)^2+2(l+1/2)^2 \} _{k,l \in \mathbb{Z}_{\ge 0}} = \{ 2(k+l+1/2)^2+2(l+1/2)^2 \} _{k,l \in \mathbb{Z}} . $$

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Notice that : $2a^2+2b^2=(a+b)^2+(a-b)^2$; so we can conclude that:

$$ 2(k+l+1/2)^2 + 2(l+1/2)^2 = (2l+k+1)^2 + k^2 \Longrightarrow \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \{ 2(k+l+1/2)^2+2(l+1/2)^2 \} _{k,l \in \mathbb{Z}} = \{ (2l+k+1)^2+k^2 \} _{k,l \in \mathbb{Z}} . $$

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Claim: $ \{ 4m^2+(2n+1)^2 \} _{m,n \in \mathbb{Z}} \subseteq \{ (2l+k+1)^2+k^2 \} _{k,l \in \mathbb{Z}} . $
Proof: Let $k:=2m$ and $l:=n-m$; one can see that: $$ (2l+k+1)^2+k^2 = (2n+1)^2+(2m)^2 . $$

Claim: $ \{ (2l+k+1)^2+k^2 \} _{k,l \in \mathbb{Z}} \subseteq \{ 4m^2+(2n+1)^2 \} _{m,n \in \mathbb{Z}} . $
Proof:

• If $k$ is even let $m:=\dfrac{k}{2}$ and $n:=l+\dfrac{k}{2}$; one can see that: $$ (2n+1)^2+(2m)^2 = (2l+k+1)^2+k^2 . $$

• If $k$ is odd let $m:=l+\dfrac{k+1}{2}$ and $n:=\dfrac{k-1}{2}$; one can see that: $$ (2m)^2+(2n+1)^2 = (2l+k+1)^2+k^2 . $$