Integer u such that c^2 ± u a square number

 
 
To find an integer which, when added or subtracted from, a square number, gives us in either case a square number

c^2 \; + \; u \; = \; p^2
c^2 \; - \; u \; = \; q^2
 
 

The integers

a \; = \; m^2 \; - \; n^2
b \; = \; 2 \, m \, n
c \; = \; m^2 \; + \; n^2

form a Pythagorean triple.
 

Here are parametric solutions:

c \; = \; m^2 \; + \; n^2
u \; = \; 4 \, m \, n \, (m^2 - n^2)
p \; = \; m^2 \; + \; 2 \, m \, n \; - \; n^2
q \; = \; -m^2 \; + \; 2 \, m \, n \; + \; n^2

(m^2 + n^2)^2 \; + \; 4 \, m \, n \, (m^2 - n^2) \; = \; (m^2 + 2 \, m \, n - n^2)^2
(m^2 + n^2)^2 \; - \; 4 \, m \, n \, (m^2 - n^2) \; = \; (-m^2 + 2 \, m \, n + n^2)^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 Integer u such that c^2 ± u a square number

  1. pipo says:

    Here are some:
    (1/2)^2 + 6 = (5/2)^2 = (7/2)^2 – 6
    Format: difference, [square1, square2, square3]
    15 [7/4, 17/4, 23/4]
    5 [31/12, 41/12, 49/12]
    14 [47/12, 65/12, 79/12]
    7 [113/120, 337/120, 463/120]
    6 [1151/140, 1201/140, 1249/140]
    13 [80929/19380, 106921/19380, 127729/19380]
    15 [112799/21896, 141121/21896, 164641/21896]
    5 [113279/1494696, 3344161/1494696, 4728001/1494696]

    pipo

  2. pipo says:

    yep, I did

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