# How can I compute a 3 by 3 unimodular matrix which produces 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

$$\begin{pmatrix} a_n \\ b_n \\ c_n \\ \end{pmatrix}= {\begin{pmatrix} 63 & 104 & −68 \\ 64 & 104 & −67 \\ 80 & 131 & −85 \\ \end{pmatrix}}^n \begin{pmatrix} 1 \\ 2 \\ 2 \\ \end{pmatrix}$$

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 ). This is the only Ramanujan-Hirschhorn matrix that I know of. I want to find at least three other Ramanujan-Hirschhorn matrices and all of them must necessarily be unimodular matrices. Can anyone show me how can I derive or compute another Ramanujan-Hirschhorn matrix different from the one given above ? Does anyone know any other Ramanujan-Hirschhorn matrix ? Can anyone help me ?

• Are there any other known families of solutions to $x^3+y^3=z^3\pm1$? – Gerry Myerson Nov 15 '19 at 23:15
• What is a "Fermat near miss"? – Rodrigo de Azevedo Nov 16 '19 at 11:41
• @GerryMyerson I found a preprint where Hirschhorn points out the Ramanujan began with an identity of a certain type. It turns out such an identity comes from a positive integer solution to $a^3 + b^3 + c^3 = d^3$ with $a+d = b+c.$ The one used was $(3,5,4,6).$ The only other choice, order $(3,4,5,6),$ fails. I do not believe that other "Ramanujan-Hirschhorn" matrices are possible. – Will Jagy Nov 16 '19 at 15:06
• see if it fits $$(3x^2 + 5 xy - 5 y^2)^3 +(4x^2 - 4xy+6y^2)^3 +(5x^2-5xy-3y^2)^3 = ( 6x^2 - 4xy+4y^2)^3$$ If we interchange the $4$ and $5$ it no longer works – Will Jagy Nov 16 '19 at 15:09
• Related discussion at math.stackexchange.com/questions/2078792/… (with contribution by @WillJagy). – Gerry Myerson Nov 16 '19 at 22:40

I don't expect any more of them. Hirschorn relies on a specific identity involving binary quadratic forms. The coefficients are parametrized by quadruples of positive integers $$(a,b,c,d)$$ such that both $$a^3 + b^3 + c^3 = d^3 \; \; \; \mbox{AND} \; \; \; a+d = b+c$$ I find just the one.

6  a: 186  b: 236  c: 313  d: 369   a + d - (b+c) :  6
6  a: 138  b: 145  c: 375  d: 388   a + d - (b+c) :  6
6  a: 108  b: 142  c: 165  d: 205   a + d - (b+c) :  6
4  a: 38  b: 43  c: 66  d: 75   a + d - (b+c) :  4
4  a: 29  b: 34  c: 44  d: 53   a + d - (b+c) :  4
4  a: 200  b: 239  c: 388  d: 431   a + d - (b+c) :  4
2  a: 46  b: 47  c: 148  d: 151   a + d - (b+c) :  2
2  a: 34  b: 39  c: 65  d: 72   a + d - (b+c) :  2
2  a: 31  b: 33  c: 72  d: 76   a + d - (b+c) :  2
2  a: 229  b: 270  c: 474  d: 517   a + d - (b+c) :  2
2  a: 145  b: 179  c: 267  d: 303   a + d - (b+c) :  2
0  a: 3  b: 4  c: 5  d: 6   a + d - (b+c) :  0
-2  a: 56  b: 61  c: 210  d: 213   a + d - (b+c) :  -2
-2  a: 53  b: 58  c: 194  d: 197   a + d - (b+c) :  -2
-2  a: 44  b: 51  c: 118  d: 123   a + d - (b+c) :  -2
-2  a: 38  b: 48  c: 79  d: 87   a + d - (b+c) :  -2
-2  a: 149  b: 166  c: 411  d: 426   a + d - (b+c) :  -2
-2  a: 134  b: 151  c: 353  d: 368   a + d - (b+c) :  -2
-2  a: 11  b: 15  c: 27  d: 29   a + d - (b+c) :  -2
-2  a: 116  b: 121  c: 607  d: 610   a + d - (b+c) :  -2
-4  a: 88  b: 95  c: 412  d: 415   a + d - (b+c) :  -4
-4  a: 7  b: 14  c: 17  d: 20   a + d - (b+c) :  -4
-4  a: 373  b: 552  c: 558  d: 733   a + d - (b+c) :  -4
-4  a: 25  b: 31  c: 86  d: 88   a + d - (b+c) :  -4
-4  a: 1  b: 6  c: 8  d: 9   a + d - (b+c) :  -4
-4  a: 16  b: 23  c: 41  d: 44   a + d - (b+c) :  -4
-4  a: 160  b: 191  c: 356  d: 383   a + d - (b+c) :  -4
-6  a: 57  b: 68  c: 180  d: 185   a + d - (b+c) :  -6
-6  a: 45  b: 69  c: 79  d: 97   a + d - (b+c) :  -6
-6  a: 45  b: 53  c: 199  d: 201   a + d - (b+c) :  -6