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

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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.

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