## Integers of the form (x^2 – 12)/y and (y^2 – 12)/x, x ≠ y

Find two distinct positive integers   $x$   and   $y$   such that

$(x, \; y) \; = \; 1$

Each of   $(x^2 - 12)/y$   and   $(y^2 - 12)/x$   is an integer.

Here are the first few solutions:

$(x, \; y) \; = \; (1, \; 11)$
$(x, \; y) \; = \; (11, \; 109)$
$(x, \; y) \; = \; (109, \; 1079)$
$(x, \; y) \; = \; (1079, \; 10681)$
$(x, \; y) \; = \; (10681, \; 105731)$
$(x, \; y) \; = \; (105731, \; 1046629)$
$(x, \; y) \; = \; (1046629, \; 10360559)$
$(x, \; y) \; = \; (10360559, \; 102558961)$
$(x, \; y) \; = \; (102558961, \; 1015229051)$
$(x, \; y) \; = \; (1015229051, \; 10049731549)$

For all members of the sequence,   $6 \, a^2 + 3$   is a square

$a = 1$   ……………   $6(1^2) + 3 = 3^2$
$a = 11$   …………..   $6(11^2) + 3 = 27^2$
$a = 109$   ………….   $6(109^2) + 3 = 267^2$
$a = 1079$   …………   $6(1079^2) + 3 = 2643^2$
$a = 10681$   ………..   $6(10681^2) + 3 = 26163^2$
$a = 105731$   ……….   $6(105731^2) + 3 = 258987^2$
$a = 1046629$   ………   $6(1046629^2) + 3 = 2563707^2$
$a = 10360559$   ……..   $6(10360559^2) + 3 = 25378083^2$
$a = 102558961$   …….   $6(102558961^2) + 3 = 2486793147^2$
$a = 1015229051$   ……   $6(1015229051^2) + 3 = 2486793147^2$
$a = 10049731549$   …..   $6(10049731549^2) + 3 = 24616714347^2$

Paul answered correctly to the question:
Is there an infinite numbers of pairs   $(x, \; y)$ ?

The square of each member of the sequence is a Centered 12-gonal numbers,
also known as, Star numbers.
that is,   $S_{n} \; = \; 6 \, n \, (n - 1) \; + \; 1$

$S_1 \; = \; 1^2$
$S_5 \; = \; 11^2$
$S_{45} \; = \; 109^2$
$S_{441} \; = \; 1079^2$
$S_{4361} \; = \; 10681^2$
$S_{43165} \; = \; 105731^2$
$S_{427285} \; = \; 1046629^2$
$S_{4229681} \; = \; 10360559^2$
$S_{41869521} \; = \; 102558961^2$
$S_{414465525} \; = \; 1015229051^2$
$S_{4102785725} \; = \; 10049731549^2$

math grad - Interest: Number theory
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### 3 Responses to Integers of the form (x^2 – 12)/y and (y^2 – 12)/x, x ≠ y

1. Paul says:

I do believe there is another pair sequence between those given.
There is a recurrence of {10, -1} from one to the next and so there will be an infinite number of pairs.

(x, y) = (1, 11)
(1^2 – 12)/11 = -1 ….. (11^2 – 12)/1 = 109

(x, y) = (11, 109)
(11^2 – 12)/109 = 1 ….. (109^2 – 12)/11 = 1079

(x, y) = (109, 1079)
(109^2 – 12)/1079 = 11 ….. (1079^2 – 12)/109 = 10681

(x, y) = (1079, 10681)
(1079^2 – 12)/10681 = 109 ….. (10681^2 – 12)/1079 = 105731

(x, y) = (10681, 105731)
(10681^2 – 12)/105731 = 1079 ….. (105731^2 – 12)/10681 = 1046629

(x, y) = (105731, 1046629)
(105731^2 – 12)/1046629 = 10681 ….. (1046629^2 – 12)/105731 = 10360559

(x, y) = (1046629, 10360559)
(1046629^2 – 12)/10360559 = 105731 ….. (10360559^2 – 12)/1046629 = 102558961

(x, y) = (10360559, 102558961)
(10360559^2 – 12)/102558961 = 1046629 ….. (102558961^2 – 12)/10360559 = 1015229051

(x, y) = (102558961, 1015229051)
(102558961^2 – 12)/1015229051 = 10360559 ….. (1015229051^2 – 12)/102558961 = 10049731549

(x, y) = (1015229051, 10049731549)
(1015229051^2 – 12)/10049731549 = 102558961 ….. (10049731549^2 – 12)/1015229051 = 99482086439

Paul.

• benvitalis says:

You are right. I missed a sequence of numbers