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)

 
 
 
 
 

 
 
 
 
 
 
 
 

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About benvitalis

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

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

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