Pos+ Integers (x,y); x + y = a^2 and x*y = b^2

To find   $(x, \; y)$   positive integers such that

$x \; < \; y$ ,     and

$x \; + \; y \; = \; a^2$
$x \, y \; = \; b^2$

$2 \; + \; 2 \; = \; 4 \; = \; 2^2$
$2 \; \times \; 2 \; = \; 4$

$5 \; + \; 20 \; = \; 25 \; = \; 5^2$
$5 \; \times \; 20 \; = \; 100 \; = \; 10^2$

$10 \; + \; 90 \; = \; 100 \; = \; 10^2$
$10 \; \times \; 90 \; = \; 900 \; = \; 30^2$

$17 \; + \; 272 \; = \; 289 \; = \; 17^2$
$17 \; \times \; 272 \; = \; 4624 \; = \; 68^2$

$26 \; + \; 650 \; = \; 676 \; = \; 26^2$
$26 \; \times \; 650 \; = \; 16900 \; = 130^2$

Note that

The sequence for the values of   x :    2,   5,   10,   17,   26

and

2
5   –   2   =   3
10   –   5   =   5
17   –   10   =   7
26   –   17   =   9
…………
…………??

Does the sequence continue?

Yes, it does.

—————————————————-

Paul found:

$37 \; + \; 1332 \; = \; 1369 \; = \; 37^2$
$37 \; \times \; 1332 \; = \; 49284 \; = \; 222^2$

$50 \; + \; 2450 \; = \; 2500 \; = \; 50^2$
$50 \; \times \; 2450 \; = \; 122500 \; = \; 350^2$

$65 \; + \; 4160 \; = \; 4225 \; = \; 65^2$
$65 \; \times \; 4160 \; = \; 270400 \; = \; 520^2$

$82 \; + \; 6642 \; = \; 6724 \; = \; 82^2$
$82 \; \times \; 6642 \; = \; 544644 \; = \; 738^2$

$101 \; + \; 10100 \; = \; 10201 \; = \; 101^2$
$101 \; \times \; 10100 \; = \; 1020100 \; = \; 1010^2$

$122 \; + \; 14762 \; = \; 14884 \; = \; 122^2$
$122 \; \times \; 14762 \; = \; 1800964 \; = \; 1342^2$

$145 \; + \; 20880 \; = \; 21025 \; = \; 145^2$
$145 \; \times \; 20880 \; = \; 3027600 \; = \; 1740^2$

$170 \; + \; 28730 \; = \; 28900 \; = \; 170^2$
$170 \; \times \; 28730 \; = \; 4884100 \; = \; 2210^2$

$197 \; + \; 38612 \; = \; 38809 \; = \; 197^2$
$197 \; \times \; 38612 \; = \; 7606564 \; = \; 2758^2$

$226 \; + \; 50850 \; = \; 51076 \; = \; 226^2$
$226 \; \times \; 50850 \; = \; 11492100 \; = \; 3390^2$

$37 \; - \; 26 \; = \; 11$
$50 \; - \; 37 \; = \; 13$
$65 \; - \; 50 \; = \; 15$
$82 \; - \; 65 \; = \; 17$
$101 \; - \; 82 \; = \; 19$
$122 \; - \; 101 \; = \; 21$
$145 \; - \; 122 \; = \; 23$
$170 \; - \; 145 \; = \; 25$
$197 \; - \; 170 \; = \; 27$
$226 \; - \; 197 \; = \; 29$

The generating functions for x and y are:
$x \; = \; n^2 \; + \; 1$
$y \; = \; n^2 \,(n^2+1)$

$x \; + \; y \; = \; (1 + n^2)^2$
$x \, y \; = \; (n + n^3)^2$

math grad - Interest: Number theory
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One Response to Pos+ Integers (x,y); x + y = a^2 and x*y = b^2

1. paul says:

Yes it does. The generating functions for x and y are:-
n^2+1 and n^2(n^2+1)
These then form the following generating functions for the squares of
x + y = (1 + n^2)^2 and
x y = (n + n^3)^2

Here are the next 10 in the sequence format {x, y, a^2, b^2}

37, 1332, 1369, 49284
50, 2450, 2500, 122500
65, 4160, 4225, 270400
82, 6642, 6724, 544644
101, 10100, 10201, 1020100
122, 14762, 14884, 1800964
145, 20880, 21025, 3027600
170, 28730, 28900, 4884100
197, 38612, 38809, 7606564
226, 50850, 51076, 11492100

Paul.