## Solve in rational numbers the system of equations — Part 1

(1)

$x^2 \; + \; x \;= \; A^2$
$x^2 \; - \; x \; = \; B^2$

(2)

$x^2 \; + \; 2 \, x \; = \; A^2$
$x^2 \; - \; 2 \, x \; = \; B^2$

Solution to   (1) :

Take any set of three squares in arithmetic progression

Let’s take   $a^2, b^2, c^2$   for the three squares, and let the common difference be   $d$.

$b^2 \; - \; a^2 \; = \; c^2 \; - \; b^2 \; = \; d$

The solution is obtained by giving   $x$   the value   $( \,b^2 \,)/d$

$x^2 \; + \; x \; = \; (b^2/d)^2 + b^2/d \; = \; (b^2 (b^2 + d))/d^2 \; = \; ( \,b^2 c^2 \,)/d^2$

and

$x^2 \; - \; x \; = \; (b^2/d)^2 - b^2/d \; = \; (b^2 (b^2 - d))/d^2 \; = \; ( \,b^2 a^2 \,)/d^2$

Here’s an example :

$a^2 = 1^2$,      $b^2 = 5^2$,      $c^2 = 7^2$

$d \; = \; 5^2 \; - \; 1^2 \; = \; 7^2 \; - \; 5^2 \; = \; 24$

$x \; = \; b^2/d \; = \; ( \,5^2 \,)/24$

$x^2 \; + \; x \; = \; (b^2 c^2)/d^2 \; = \; (5^2 \cdot 7^2)/24^2 \; = \; ( \,35/24 \,)^2$

$x^2 \; - \; x \; = \; (b^2 a^2)/d^2 \; = \; (5^2 \cdot 1^2)/24^2 \; = \; ( \,5/24 \,)^2$

Solution of (2) :   The method is similar to that in (1)

math grad - Interest: Number theory
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### 2 Responses to Solve in rational numbers the system of equations — Part 1

1. pipo says:

The first pair (p,x) where x*(x ± p) both are squares is (24,25).
Other pairs are of course (48, 50), (72,75) etc, n times the primitive solution.
Other primitive pairs (p,x) are:
(120, 169)
(240, 289)
(336, 625)
(720, 1681)
(840, 841)
(840, 1369)
(1320, 3721)
(2016, 4225)
(2184, 7225)
(2520, 2809)
(3696, 4225)
(5280, 5329)
(5544, 7225)
(6240, 7921)
(9360, 9409)
pipo

• benvitalis says:

Here’s my solution to part (1)