## 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:

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

$(m^2 + n^2)^2 \; + \; 4 \, m \, n \, (m^2 - n^2) \; = \; (m^2 + 2 \, m \, n - n^2)^2$
$(m^2 + n^2)^2 \; - \; 4 \, m \, n \, (m^2 - n^2) \; = \; (-m^2 + 2 \, m \, n + n^2)^2$

math grad - Interest: Number theory
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### 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

• benvitalis says:

Nice. You used sets of 3 squares in arithmetic progression
23^2 – 17^2 = 17^2 – 7^2 = 240 = 4^2 * 15
getting 23/4, 17/4, 7/4

2. pipo says:

yep, I did