An Isomorphism Problem in Z ^ 2

We consider Euclidean distance graphs with vertex set Q2 or Z2 and address the possibility or impossibility of finding isomorphisms between such graphs. It is observed that for any distances d1, d2 the non-trivial (that is, having non-empty edge set) distance graphs G(Q2, d1) and G(Q2, d2) are isomorphic. Ultimately it is shown that for distinct primes p1, p2 the non-trivial distance graphs G(Z2, √ p1) and G(Z2, √ p2) are not isomorphic. We conclude with a few additional questions related to this work.

Note also that Euler's characterization implies that if a pair of integers x, y are each representable as the sum of two integer squares, then their product xy is also representable as a sum of two integer squares.A relevant extension is the fact that if two non-zero rational numbers q 1 , q 2 are each representable as a sum of two rational squares, then both their product and their ratio are representable as a sum of two rational squares.To see this, observe that if q 1 = a 1 2 + b 1 2 and q 2 = a 2 2 + b 2 2 for a 1 , a 2 , b 1 , b 2 ∈ Q, then q 1 q 2 = (a 1 a 2 − b 1 b 2 ) 2 + (a 1 b 2 + b 1 a 2 ) 2 and q 1 q 2 = ( a 1 a 2 −b 1 b 2 q 2 ) 2 + ( a 1 b 2 −b 1 a 2 q 2 ) 2 .This gives rise to the following theorem.
Furthermore, this transformation is a bijection as the matrix a b b −a is invertible.Thus So of course all non-trivial Euclidean distance graphs with vertex set Q 2 have the same chromatic number.They are all just the same graph.This observation is not noteworthy in itself.It is basically just a classical argument presented in a new setting.However, we are now wondering what happens when we ask a similar question about Euclidean distance graphs with vertex set Z 2 .In other words, for distinct positive integers z 1 , z 2 with , there are two immediate conditions which would result in the graphs being non ).These issues, however, can be addressed with elementary arguments along with an appeal to classical number theory.
To begin, note that for any graph is simply the number of different ordered signed representations of z as a sum of two integer squares.This number is well-known (see [3]), and extends Lemma 2.1 from the previous section.
with each p i being a distinct prime congruent to 1 (mod 4) and each q j being a distinct prime congruent to 3 (mod 4).Then the number of different ordered signed representations of z as a sum of two integer squares is 4(β 1 + 1)(β 2 + 1) Proof.We first show that the vector 0, 1 can be realized as a sum of Z 2 vectors of length √ z.By Lemma 3.1, there exist integers a, b such that a 2 + b 2 = z and gcd(a, b) = 1.As z is odd, we have that exactly one of a, b is even.So assume a is even and for the sake of clarity, assume a and b are non-negative.As gcd(a, b) = 1, let s, t be non-negative integers such that sa The fact that 0, 1 can be expressed as a sum of Z 2 vectors of length √ z implies that 1, 0 , −1, 0 , and 0, −1 can be as well.As any vector in Z 2 can be expressed as a combination of the vectors 0, 1 , 1, 0 , −1, 0 , and 0, −1 , we have established that G(Z We now mimic the proof of Theorem 3.2, except this time we create the vector 2, 0 as a sum of Z 2 vectors of length By similar arguments, 0, 2 , −2, 0 , and 0, −2 can be created as a sum of the desired vectors as well.Furthermore, we may then add some number of the vectors −2, 0 and 0, −2 to a, b to create 1, 1 .Now observing that any Z 2 vector can be obtained by beginning with either 0, 0 or 1, 0 and adding to that vector some number of the vectors 2, 0 , 0, 2 , −2, 0 , 0, −2 , or 1, 1 , we have that k(Z 2 , √ 2z) = 2.
Regarding Lemma 3.1, note that for any integers a, b and prime q ≡ 3 (mod 4), q 2 |(a 2 +b 2 ) implies that q|a and q|b.Thus in the graph G(Z 2 , zq 2 ) where z is any positive integer, vertices (x 1 , y 1 ) and (x 2 , y 2 ) are in the same component only if x 1 ≡ x 2 (mod q) and y 1 ≡ y 2 (mod q).Extending this fact in conjunction with the previous two theorems, we obtain the following result.
with each p i being a distinct prime congruent to 1 (mod 4) and each q j being a distinct prime congruent to 3 (mod 4).Then In light of the preceding work, it appears that a thorough analysis of the question posed at the end of Section 2 can be reached by only considering graphs of the form G(Z 2 , √ z) where the prime factorization of z consists solely of factors congruent to 1 (mod 4).We begin this analysis below, first making note of a graph invariant that we will use along the way.
Lemma 3.5.Let G, H be graphs and let f : Proof.We need only note that there is a bijection φ between the set of all closed walks in G containing v and the set of all closed walks in H containing f (v).For any closed walk Theorem 3.6.Let p, q be distinct primes congruent to 1 (mod 4).
Proof.For the sake of brevity, let for any vertex u we need only determine the number of distinct ordered collections of l vectors, each having integer entries and length √ p, whose sum is the zero vector.We calculate W (G 2 , v, l) in the same fashion except requiring that each vector has length √ q.To establish that G 1 G 2 , it suffices to show that for some In the interest of economy of words, we will refer to an ordered collection of vectors that sum to the zero vector as just being a "collection".Furthermore, we will classify each collection as being trivial or non-trivial.A trivial collection is one where for each integer z, the number of times z appears as an x-component entry in the collection is equal to the number of times −z appears as an x-component entry in the collection and the number of times z appears as an y-component entry in the collection is equal to the number of times −z appears as an y-component entry in the collection.A non-trivial collection is any that does not have this property.
Let a, b, c, d ∈ Z + such that a 2 + b 2 = p and c 2 + d 2 = q.By Lemma 3.1, the vectors a, b and c, d , along with those formed by permuting entries or replacing an entry with its negative, are the only vectors of length √ p and √ q respectively.It follows that for any l ∈ Z + , the number of distinct trivial collections of l vectors of length √ p is equal to the of distinct trivial collections of vectors of length √ q.
We now determine the minimum value l such that a non-trivial collection of l vectors of length √ p actually exists.First note that if such a non-trivial collection is minimum, then either the set of all x-entries in the collection cannot contain both a and −a or b and −b or the set of all y-entries in the collection cannot contain both a and −a or b and −b.Without loss of generality, we may assume a minimum collection contains only a and −b as x-entries of the vectors in that collection.Therefore, a minimum collection must be of length s + t where s, t ∈ Z + and sa − tb = 0.As gcd(a, b) = 1, we have that (s + t) is a multiple of (a + b).However, s + t = a + b as the y-component entries of this collection would then consist of b copies of ±b and a copies of ±a.As exactly one of a, b is even, it would then be impossible for the y-entry of the sum of this collection of vectors to be zero.

Further Work
It is our guess that there do not exist distinct z 1 , z 2 ∈ Z + such that the non-trivial distance graphs G(Z 2 , √ z 1 ) and G(Z 2 , √ z 2 ) are isomorphic.In light of the work done in the previous section, we may completely resolve this matter by considering pairs of graphs of the form where z 1 , z 2 are composite, have prime factorizations consisting solely of factors congruent to 1 (mod 4), and if α 1 , ..., α m and β 1 , ..., β n are the multiplicities of the prime factors of z 1 and z 2 respectively, then (α ).So at least that narrows down our focus.However, there is another possibly interesting direction we can take this question.A taste of this can be seen in the following example.
Consider the graphs G(Z 2 , √ 5) and G(Z 2 , √ p) where p is any prime greater than 5.By Theorem 3.6, we know G(Z 2 , √ 5) G(Z 2 , √ p).We are now wondering if this can be shown directly by constructing a graph H that is a subgraph of one of these graphs but not of the other.Well, it just so happens that it can.
Proof.Suppose H does appear as a subgraph of G(Z 2 , √ p).Label the vectors corresponding to the edges of H as in Figure 2.
Note that we are depicting H with the same visual representation as it has when appearing as a subgraph of G(Z 2 , √ 5).This certainly does not have to be the case, but we feel it aids in grasping the argument that follows.Note also that once we have labeled u 1 , u 2 , u 3 , v 1 , v 2 , v 3 as in Figure 2, our hand is forced for labeling the remaining edges with their corresponding vectors.This is done in Figure 3.
We now observe that u 1 + u 2 + u 3 + v 1 + v 2 + v 3 = 0, but that none of the following sums of two vectors equal the zero vector: and v 1 = v 3 .Let p = a 2 +b 2 for a, b ∈ Z + and note that a+b > 3.In regards to the work done in the proof of Theorem 3.6, we have that T = {u 1 , u 2 , u 3 , v 1 , v 2 , v 3 } is a trivial collection of six vectors of length √ p that sum to the zero vector.We need only consider the following two cases.
Case 1 Suppose the x-entries of the six vectors of consist of two copies of a, two copies of −a, and one copy each of and −b.
By the above observations, for any vector v ∈ T , u 1 + v = 0 and v 1 + v = 0.As the two vectors with x-entries of b or −b must be additive inverses, it follows that neither of those vectors can be equal to u 1 or v 1 .Also in accordance with the previous observations, it follows that one of the two vectors with x-entries of b or −b is equal to u i for i ∈ {2, 3} if and only if the other of the two vectors is equal to v i .
Assuming the two vectors with x-entries of b or −b are equal to u 2 and v 2 , since u 1 = u 3 and u 1 + u 3 = 0, we have that u 1 and u 3 agree on exactly one of their two component entries.This implies that either u 1 + v 1 = 0 or u 3 + v 1 = 0, contradicting our previous observations.
Assuming the two vectors with x-entries of b or −b are equal to u 3 and v 3 , since u 1 = u 2 and u 1 + u 2 = 0, we have that u 1 and u 2 agree on exactly one of their two component entries.This implies that either u 1 + v 1 = 0 or u 2 + v 1 = 0, again contradicting our previous observations.Case 2 Suppose the x-entries of the six vectors of T consist of three copies of a and three copies of −a.
We have that u 1 + v 1 = 0.If u 1 = v 1 , we would then have that each of u 2 u 3 , v 2 , v 3 would be an additive inverse for one of u 1 , v 1 , contradicting our previous observations.Assuming that u 1 = v 1 and again noting that none of u 2 , u 3 , v 2 , v 3 are additive inverses of u 1 , v 1 and noting that u 1 ∈ {u 2 , u 3 } and v 1 ∈ {v 2 , v 3 }, it follows that the set {u 2 , u 3 , v 2 , v 3 } contains at most two distinct vectors, and if it contains two, then those two vectors are additive inverses of each other.As u 2 + u 3 = 0 and u 2 + v 3 = 0, it follows that u 2 = u 3 = v 3 .As v 2 + v 3 = 0 and v 2 + u 3 = 0, it follows that v 2 = v 3 = u 3 .Thus u 2 = u 3 = v 2 = v 3 , contradicting the fact that the x-entries of T sum to 0.
Admittedly, the proof of Theorem 4.1 is a bit tedious and awkward.However, it does serve as a solution to a specific instance of the following general question, which we feel is of interest and offers an avenue for future work.
Question 1 Given non-isomorphic graphs G(Z 2 , d 1 ) and G(Z 2 , d 2 ), if possible, construct a finite graph H which appears as a subgraph of exactly one of the two given graphs.
Better yet, answer the following.

Acknowledgements
The author thanks Peter D. Johnson of Auburn University and Joseph Chaffee, currently a at large, for their input at various stages of the creation of this article.
Now noting that a −b, a + a −b, −a + b a, b + b a, −b = 0, 0 , we have established that a minimum non-trivial collection of l vectors of length √ p has l = 2(a + b).By a similar argument, a minimum non-trivial collection of l vectors of length√ q has l = 2(c + d).So if a + b = c + d, we have that G(Z 2 , √ p) G(Z 2 ,√ q) and we're done.Now assume that 2(a + b) = 2(c + d) = l.Consider a non-trivial collection of l vectors of length √ p whose x-component entries consist solely of a and −b.To construct such a collection, we have 2a+2b 2b choices of how to distribute the 2b copies of a among the ordered x-entries of the vectors.Once we make such an assignment of values to the x-entries, we then have 2a a choices of how to distribute the a copies each of the terms a and −a among the ordered y-entries of the vectors.Similarly, we have 2b b choices of how to distribute the b copies each of the terms b and −b among the ordered y-entries of the vectors.Combining these facts, and observing that we may create additional non-trivial collections of l vectors of length √ p by simultaneously exchanging the x-and y-entries of each of the l vectors or by simultaneously replacing each copy of a with −a and each copy of −b with b in each vector of the collection, we obtain the fact that the number of distinct non-trivial collections of l vectors of length √ p is given as 4 2a+2b 2b 2a a 2b b , which may be simplified to 4 l! (a!b!) 2 .Moreover, the number of distinct non-trivial collections of l vectors of length √ q is equal to 4 l! (c!d!) 2 .Let n = a + b = c + d and observe that a, b, c, d must be distinct.Without loss of generality, assume a < c < d < b and note that this implies a ≤ n−1 2 − 1. Define a function f (x) = x!(n − x)! and consider the behavior of f (x) on the domain {1, 2, ..., n−1 2

Question 2
Given non-isomorphic graphs G(Z 2 , d 1 ) and G(Z 2 , d 2 ), construct a graph H as stipulated by Question 1 where either |V (H)| or |E(H)| is minimum.