# How can you verify that a 3 by 3 unimodular matrix generates an infinite number of Fermat near misses?

I am interested in finding 3 by 3 Ramanujan-Hirschhorn matrices.

By definition, a Ramanujan-Hirschhorn matrix is a 3 by 3 matrix which produces an infinite number of Fermat near misses.

Ramanujan in his "lost notebook" makes the amazing claim that if the integers a_n, b_n, c_n are defined by:

$$\sum_{x\ge0}{a_n x^n}=\frac{1 + 53x + 9x^2}{1 − 82x − 82x^2 + x^3}$$

$$\sum_{x\ge0}{b_n x^n}=\frac{2 - 26x - 12x^2}{1 − 82x − 82x^2 + x^3}$$

$$\sum_{x\ge0}{c_n x^n}=\frac{2 + 8x - 10x^2}{1 − 82x − 82x^2 + x^3}$$

then $$a_n^3 + b_n^3 = c_n^3 + (-1)^n$$

Two proofs of this claim and a plausible explanation of how Ramanujan may have been led to it have been given by Michael Hirschhorn (1993 -94 ). Indeed, Hirschhorn showed that the sequences {a_n} , {b_n} and {c_n} are given by $$\begin{array}{} \begin{bmatrix} a_n \\\ b_n \\\ c_n \end{bmatrix} & = {\begin{bmatrix} 63 & 104 & −68 \\\ 64 & 104 & −67 \\\ 80 & 131 & −85 \end{bmatrix}}^n & \cdot & \begin{bmatrix} 1 \\\ 2 \\\ 2 \end{bmatrix} \end{array}$$ Notice that the matrix above is unimodular and it produces an infinite number of triples of Fermat near misses when multiplied on the right by the column vector (1 2 2 ). In 2013 Tito Piezas , with the help of Mathematica's GeneratingFunction command was able to find: \begin{aligned} \sum_{n=0}^\infty a_n x^n &= \frac{-9(417-5602x+x^2)}{R_2}\\\ \sum_{n=0}^\infty b_n x^n &= \frac{8(-566-11315x+x^2)}{R_2}\\\ \sum_{n=0}^\infty c_n x^n &= \frac{-6(877+6898x+x^2)}{R_2} \end{aligned} where $$R_2 = -1+184899x-184899x^2+x^3$$ and,

$$a_n^3+b_n^3 = c_n^3 + 1$$

which is a sum of cubes identity analogous to Ramanujan's. I used a method very similar to Michael Hirschhorn's method to derive a 3 by 3 unimodular matrix associated with the above Tito Piezas sum of cubes identity. The unimodular matrix that i have found is: $$\begin{array} {} \begin{bmatrix} a_n \\\ b_n \\\ c_n \end{bmatrix} &=& {\begin{bmatrix} 156625 & 115992 & −79656 \\\ 189000 & 139969 & −96120 \\\ 219624 & 162648 & −111695 \end{bmatrix}}^n \cdot \begin{bmatrix} 3753 \\\ 4528 \\\ 5262 \end{bmatrix} \end{array}$$ As you can see the above matrix is also unimodular and according to Tito Piezas generates an infinite number of Fermat near miss triples when multiplied on the right by the column vector (3753 4528 5262 ). But how can i verify the claim that the above matrix produces an infinite number of Fermat near miss triples? Any suggestions or ideas ? More importantly, how can I derive or compute another Ramanujan-Hirschhorn matrix different from the one given above ? Is there a systematic way of deriving these matrices ? There is probably an infinite number of them !! Does anyone know any other Ramanujan-Hirschhorn matrices ?

• @GerryMyerson can you please link this question to my question last November: "How can I compute a 3 by 3 unimodular matrix which produces an infinite number of Fermat near misses?" Jan 10, 2020 at 22:11
• @GerryMyerson Can you please link this question to my question last November: "How can I compute a 3 by 3 unimodular matrix which produces an infinite number of Fermat near misses?" Jan 10, 2020 at 22:13
• @WillJagy can you please edit the matrix equation so that it looks right. can you please link this question to my question last November: "How can I compute a 3 by 3 unimodular matrix which produces an infinite number of Fermat near misses?" – Jan 10, 2020 at 22:18

Note : this answer has been simplified & improved at Jan 12, 9:16.

Here is an answer to your first question (the second one is much harder AFAIK).

Let

$$M={\begin{bmatrix} 156625 & 115992 & −79656 \\\ 189000 & 139969 & −96120 \\\ 219624 & 162648 & −111695 \end{bmatrix}},$$

$$V_0= \begin{bmatrix} 3753 \\\ 4528 \\\ 5262 \end{bmatrix}, V= \begin{bmatrix} x \\\ y \\\ z \end{bmatrix}, d\begin{pmatrix}x \\\ y \\\ z \end{pmatrix} = x^3+y^3-z^3-1$$

Then, one has the identity :

$$d(M^3V)=184899d(M^2V)-184899d(MV)+d(V) \tag{1}$$ (you can check (1) by hand or with the help of a computer if you feel lazy).

It follows from (1) that $$d(M^3V)=0$$ if $$d(V)=d(MV)=d(M^2V)=0$$. By induction, we deduce $$d(M^nV)=0$$ (for $$n\geq 0$$) whenever $$d(V)=d(MV)=d(M^2V)=0$$. But it is straightforward to see that this is the case for $$V=V_0$$, and the result follows.

Note that the coefficients appearing in (1) are exactly the coefficients of the characteristic polynomial of $$M$$ ; note also that this polynomial is a "reciprocal" polynomial.