fmn and tfmn: The Functions f and tf

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1. Area and progression difference

For nonzero rational m,n, define

f(m,n)=mn(m^2-n^2) =mn(m-n)(m+n).

This form arises from the classical parametrization of Pythagorean triples:

(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2,\qquad \frac{(m^2-n^2)(2mn)}2=f(m,n).

For m>n>0, the value f(m,n) is the area of the right triangle with legs m²−n² and 2mn. For arbitrary orders and signs of the parameters, f also records orientation.

The connection with arithmetic progressions of squares follows from the identities

\begin{aligned} (m^2+n^2)^2-4f(m,n)&=(-m^2+2mn+n^2)^2,\\ (m^2+n^2)^2+4f(m,n)&=(m^2+2mn-n^2)^2. \end{aligned}

Consequently, the three expressions

V-4f(m,n),\qquad V,\qquad V+4f(m,n), \qquad V=(m^2+n^2)^2,

are squares and form an arithmetic progression with oriented difference 4f(m,n). This identity underlies the application of the fmn construction to the eight progressions in a magic square.

2. The squarefree part

For an integer N≠0, let t(N) denote the squarefree part of N. It is the unique squarefree integer with the same sign as N for which

N=t(N)q^2,\qquad q\in\mathbb Z_{>0}.

It can be read directly from the prime factorization. If

N=\operatorname{sgn}(N)\prod_p p^{e_p},

then

t(N)=\operatorname{sgn}(N)\prod_{e_p\ \mathrm{odd}}p, \qquad q=\prod_p p^{\lfloor e_p/2\rfloor}.

The product t(N) contains exactly the primes occurring in N to an odd exponent; all remaining powers are absorbed by the square q². This proves existence. Uniqueness follows from unique prime factorization: the sign of t(N) is fixed by the sign of N, and the exponent of each prime in a squarefree part can only be 0 or 1, so it must equal eₚ modulo 2.

In the language of square classes, the same construction is expressed by the equivalence

t(A)=t(B) \iff \frac AB\in(\mathbb Q^\times)^2, \qquad A,B\in\mathbb Z\setminus\{0\}.

Thus t(N) is the canonical integral representative of the class of N in ℚ×/(ℚ×)². This abstract interpretation is useful in proofs, but the function t itself will simply mean the squarefree part.

Definition of the function tf

\operatorname{tf}(m,n)=t\!\left(f(m,n)\right), \qquad m,n\in\mathbb Z,\quad f(m,n)\ne0.

Thus tf(m,n) is the squarefree part of f(m,n). When f(m,n)=0, the progression is constant and tf is undefined.

3. tf(m,n) as a congruent number

Let m>n>0 be integers and write

f(m,n)=Tq^2,\qquad T=\operatorname{tf}(m,n)>0.

Divide the sides of the corresponding Pythagorean triangle by q:

A=\frac{m^2-n^2}{q},\qquad B=\frac{2mn}{q},\qquad H=\frac{m^2+n^2}{q}.

This gives a rational right triangle of area

\frac{AB}{2}=\frac{mn(m^2-n^2)}{q^2}=T.

Therefore a positive value of tf(m,n) is the squarefree part of the area and is also a congruent number. Distinct pairs (m,n) may have the same squarefree part and hence determine the same T.

4. Criteria for equality of tf values

Theorem

Let Fᵢ=f(mᵢ,nᵢ) be two nonzero integral values of f and let G=gcd(|F₁|,|F₂|). The following conditions are equivalent:

$t(F_1)=t(F_2)$ ;
$F_1/F_2$ is a square in ℚ;
$F_1F_2$ is a positive perfect square;
F₁,F₂ have the same sign, and $|F_1|/G$ and $|F_2|/G$ are perfect squares.

Proof

Write Fᵢ=Tᵢqᵢ², where Tᵢ=t(Fᵢ) is the squarefree part of Fᵢ. The quotient F₁/F₂ is a square exactly when the sign and the parity of every prime exponent agree, that is, exactly when T₁=T₂. The product F₁F₂ is a positive square under the same condition.

If T₁=T₂=T, then G=|T|gcd(q₁,q₂)², so both quotients |Fᵢ|/G are squares. Conversely, if those two quotients are squares and the signs agree, their quotient F₁/F₂ is a rational square. Hence all four conditions are equivalent.

5. Matching the scales of progressions

Multiplying all three terms of a progression of squares by λ² preserves squarehood and multiplies its difference by λ². Thus two parametrized progressions with values F₁=f(m₁,n₁) and F₂=f(m₂,n₂) admit a common oriented difference exactly when there exist nonzero α,β∈ℚ such that

4\alpha^2F_1=4\beta^2F_2.

After cancellation, this equality is precisely the second condition of the preceding theorem. Therefore,

\operatorname{tf}(m_1,n_1)=\operatorname{tf}(m_2,n_2) \iff \text{the progressions scale to a common oriented difference}.

Constructively, if F₁=Tu² and F₂=Tv², multiply the first progression by v² and the second by u². Their differences both become 4Tu²v². This is the arithmetic content of equality between tf values.

6. Parameter symmetries

Some equalities between tf values arise from parameter substitutions that preserve the square class or merely reverse the orientation of the progression:

Substitution	Effect on f	Meaning
$(m,n)\mapsto(\lambda m,\lambda n)$	$f\mapsto\lambda^4f$	common parameter scale
$(m,n)\mapsto(-m,-n)$	$f\mapsto f$	simultaneous sign change
$(m,n)\mapsto(n,m)$	$f\mapsto-f$	reversal of the progression
$(m,n)\mapsto(m,-n)$	$f\mapsto-f$	reversal of the progression
$(m,n)\mapsto(m+n,m-n)$	$f\mapsto4f$	the same normalized progression

The last row follows immediately from the factorization f(a,b)=ab(a+b)(a−b):

\begin{aligned} f(m+n,m-n) &=(m+n)(m-n)\cdot2m\cdot2n\\ &=4f(m,n). \end{aligned}

The middle square V=(m²+n²)² is also multiplied by 4 under the same substitution. Hence the ratio 4f/V, namely dir, is unchanged. Since the factor 4 is a square, the value of tf is preserved as well. When unoriented progressions are classified, opposite signs of tf are identified as well.

7. The self-recurrence of tf

A progression of squares itself supplies a new parameter pair. Set

V=(m^2+n^2)^2,\qquad D=4f(m,n).

If r=−m²+2mn+n², s=m²+n², and w=m²+2mn−n², then V=s², V−D=r², and V+D=w². Therefore

\begin{aligned} f(V,D) &=VD(V^2-D^2)\\ &=s^2\cdot4f(m,n)\cdot r^2w^2\\ &=f(m,n)(2srw)^2. \end{aligned}

The quotient f(V,D)/f(m,n) is a square, and the square-class theorem gives the identity

\operatorname{tf}(m,n) = \operatorname{tf}\!\left((m^2+n^2)^2,\;4f(m,n)\right).

8. The elliptic curve of a fixed tf value

For T=tf(m,n)>0 and f(m,n)=Tq², the rational triangle from Section 3 corresponds to a point on the congruent-number elliptic curve

E_T:\quad y^2=x^3-T^2x.

In terms of m,n, this point has the particularly simple form

P_{m,n}= \left( \frac{Tm}{n}, \frac{T^2q}{n^2} \right).

Substitution of f(m,n)=Tq² directly verifies the curve equation. Conversely, a point (x,y) with y≠0 gives the rational right triangle

A=\frac{x^2-T^2}{y},\qquad B=\frac{2Tx}{y},\qquad H=\frac{x^2+T^2}{y},

for which A²+B²=H² and AB/2=T. Parametrizing this triangle recovers the rational ratio m:n. Thus the pairs with fixed tf(m,n)=T parametrize rational points on one and the same curve E_T, modulo the symmetries described above.

The self-recurrence of the preceding section has a standard meaning on this curve. The duplication formula gives

x(2P_{m,n}) = T\,\frac{(m^2+n^2)^2}{4f(m,n)} =T\,\frac VD.

This is the x-coordinate of the point constructed from the new pair (V,D); the choice of orientation determines the sign of y. Thus the self-recurrence is point doubling rather than an independent source of a second point on E_T.

9. From the tfmn method to 6/9 configurations

Equality of tf values solves the problem of matching the differences of two progressions of squares. To place both progressions in one magic square, their middle terms, shared entries, and linear coordinates E,x,y must also be compatible. Those conditions depend on the positional type of the configuration.

For 6/9 classes containing two parallel progressions, the tf value is the natural arithmetic coordinate of the tfmn method: one first chooses two points on the same curve E_T and then solves the remaining equations that place them in the general form of a magic square. In the classes studied so far, the additional placement equations lead to conics and elliptic surfaces associated with particular families.

The early F1–F8 families →General theory of 4/9 and 5/9 patterns