representation of integers by equivalent integral binary quadratic forms

Theorem 1.

If F,G are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath integral binary quadratic forms, then F and G represent the same set of integers.


Write G(x,y)=F(αx+βy,γx+δy) where


Then m=G(r,s)m=F(αr+βs,γr+δs), so if G represents m, so does F. Since the matrix has determinantMathworldPlanetmath 1, it is invertiblePlanetmathPlanetmathPlanetmath and its inversePlanetmathPlanetmathPlanetmath is another integer matrix, so the reverse statement follows as well. ∎

Lemma 2.

F properly represents an integer m if and only if F is properly equivalent to a form mx2+Bxy+Cy2.


: It is obvious by the above that F represents m; the problem is to show that it represents m properly. Write G(x,y)=mx2+Bxy+Cy2; then G(x,y)=F(αx+βy,γx+δy), where αδ-βγ=1. Then m=G(1,0)=F(α,γ). But clearly (α,γ)=1 since otherwise we cannot have αδ-βγ=1. So F represents m properly.
: Write F(p,q)=m, where (p,q)=1. Since (p,q)=1, we can find integers r,s such that ps-qr=1, and then


Definition 1.

If F is a binary quadratic form, its discriminantMathworldPlanetmathPlanetmath, Δ(F) is b2-4ac.

Note that Δ(F) is always either congruentMathworldPlanetmath to 0 or 1 mod 4, and that b is even (odd) exactly when Δ(F)0(1)(mod4).

Theorem 3.

If F,G are equivalent integral quadratic formsMathworldPlanetmath, then Δ(F)=Δ(G).


For any form F, define



2F(x,y)=(x y)MF(xy)

Note further that Δ(F)=-det(MF).

Now in our particular case, if G(x,y)=F(αx+βy,γx+δy), then

2G(x,y)=(αx+βy γx+δy)MF(αx+βyγx+δy)=(x y)(αγβδ)MF(αβγδ)(xy)



But Δ(F)=-det(MF), so since det(αγβδ)=det(αβγδ)=±1,


Note that this proof shows that applying a set of transformations amount to multiplying by the transform matrix on the left and its transposeMathworldPlanetmath on the right.

Example: In the previous example, note that Δ(F)=1-416=-23, and Δ(G)=512-4828=2601-2624=-23.

The converseMathworldPlanetmath of this theorem is not true - that is, there are forms of the same discriminant that represent different numbers. For example, x2+5y2 and 2x2+2xy+3y2 both have discriminant -20, yet the second form represents 2 while the first clearly does not. However, equivalence classesMathworldPlanetmath of forms under arbitrary (proper or improper) equivalence represent disjoint sets of primes:

Theorem 4.

Let p be an odd prime. Suppose F,G both represent p and Δ(F)=Δ(G). Then F and G are equivalent (but perhaps not properly equivalent).


Since p is prime, F obviously represents p properly. So Fpx2+bxy+cy2. Note that the transformation (x,y)(x+dy,y) results in a form whose middle term is 2pd+b, so by an appropriate choice of d we can arrange that -p<bp. Similarly, Gpx2+bxy+cy2 with -p<bp. Note also that since b2-4pc=b2-4pc, it follows that bb(2) (i.e. b,b have the same parity).

Since Δ(F)=Δ(G), we see that b2-4pc=b2-4pcb2b2(p)b±b(p), so b=±b+kp for some k. Since b,b have the same parity and p is odd, k is even; since -p<b,bp, k=0 (since otherwise b,b would be separated by at least 2p, which is impossible).

We are left with two cases. If b=b, then Δ(F)=Δ(G) implies that c=c and hence FG. If b=-b, then again Δ(F)=Δ(G) implies that c=c. Then F and G are equivalent via the transformation (x,y)(x,-y). ∎

Note that F(x,y)=ax2+bxy+cy2 and G(x,y)=ax2-bxy+cy2 are always improperly equivalent via the transformation (x,y)(x,-y). They are sometimes properly equivalent, and sometimes not. For example, 2x2+2xy+3y2 and 2x2-2xy+3y2 are properly equivalent while 3x2+2xy+5y2 and 3x2-2xy+5y2 are not. (See the article on reduced integral binary quadratic forms for details).

In summary, we have proved the following:

F,G equivalent F,G represent the same set of integers 
F,G equivalent Δ(F)=Δ(G)
Δ(F)=Δ(G) and F,G both represent some odd prime pF and G are equivalent

We conclude with the following lemma and corollary, which provide concrete criteria for when an integer is representable by a class of forms.

Lemma 5.

If D0,1(4) is an integer, and m is an odd integer relatively prime to D, then m is properly represented by a primitive form of discriminant D if and only if D is a quadratic residueMathworldPlanetmath modm.


If F(x,y) properly represents m, then by the preceding lemma, we may assume that F(x,y)=mx2+bxy+cy2. Then D=b2-4mc, being the discriminant of F, so that Db2(D). Conversely, if Db2(D), we may assume Db(2) (if they have different parities, replace b by b+m; since m is odd, the condition now holds and D(b+m)2(D) as well). Since D0,1(4), it follows that Db2(4) and thus Db2(4m). Hence D=b2-4mc for some integer c. But then mx2+bxy+cy2 represents m and has discriminant D; it is primitive since gcd(m,b)=gcd(m,D)=1. ∎

Corollary 6.

Let n be an integer, and p an odd prime not dividing n. Then (-np)=1 if and only if p is represented by a primitive form of discriminant -4n.


By the preceding lemma, p is represented by a primitive form of discriminant -4n if and only if


Title representation of integers by equivalent integral binary quadratic forms
Canonical name RepresentationOfIntegersByEquivalentIntegralBinaryQuadraticForms
Date of creation 2013-03-22 19:18:48
Last modified on 2013-03-22 19:18:48
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 7
Author rm50 (10146)
Entry type Topic
Classification msc 11E12
Classification msc 11E16
Related topic integralbinaryquadraticforms