## When a*b + c*d, a*d – b*c are squares — Part 1

Can you find positive integers   $a, b, c, d$   so that the expressions

$a \, b \; + \; c \, d$
$a \, d \; - \; b \, c$

Note that

$(a^2 + c^2) \,(b^2 + d^2) \; = \; (a \, b + c \, d)^2 \; + \; (a \, d - b \, c)^2$

I want to produce

$(a^2 + c^2) \,(b^2 + d^2) \; = \; x^4 \; + \; y^4$

For example,

$(4\times 5) \; + \; (11\times 16) \; = \; 14^2$ …….. $(4\times 16) \; - \; (5\times 11) \; = \; 3^2$

$(4^2 + 11^2) \,(5^2 + 16^2) \; = \; 14^4 \; + \; 3^4$

$(3\times 4) \; + \; (8\times 11) \; = \; 10^2$ ……… $(3\times 11) \; - \; (4\times 8) \; = \; 1^2$

$(3^2 + 8^2) \,(4^2 + 11^2) \; = \; 10^4 \; + \; 1^4$

$(3\times 7) \; + \; (20\times 47) \; = \; 31^2$ …….. $(3\times 47) \; - \; (7\times 20) \; = \; 1^2$

$(3^2 + 20^2) \,(7^2 + 47^2) \; = \; 31^4 \; + \; 1^4$

math grad - Interest: Number theory
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### 3 Responses to When a*b + c*d, a*d – b*c are squares — Part 1

1. paul says:

Here are a few, a < b && c < d <= 50

3 x 4 + 8 x 11 = 10^2
3 x 11 – 4 x 8 = 1^2

3 x 7 + 20 x 47 = 31^2
3 x 47 – 7 x 20 = 1^2

4 x 5 + 11 x 16 = 14^2
4 x 16 – 5 x 11 = 3^2

4 x 5 + 16 x 29 = 22^2
4 x 29 – 5 x 16 = 6^2

4 x 9 + 15 x 36 = 24^2
4 x 36 – 9 x 15 = 3^2

4 x 9 + 16 x 40 = 26^2
4 x 40 – 9 x 16 = 4^2

5 x 6 + 21 x 26 = 24^2
5 x 26 – 6 x 21 = 2^2

5 x 7 + 10 x 19 = 15^2
5 x 19 – 7 x 10 = 5^2

5 x 8 + 10 x 36 = 20^2
5 x 36 – 8 x 10 = 10^2

5 x 8 + 13 x 45 = 25^2
5 x 45 – 8 x 13 = 11^2

5 x 9 + 11 x 36 = 21^2
5 x 36 – 9 x 11 = 9^2

5 x 9 + 11 x 44 = 23^2
5 x 44 – 9 x 11 = 11^2

5 x 9 + 19 x 36 = 27^2
5 x 36 – 9 x 19 = 3^2

5 x 11 + 15 x 38 = 25^2
5 x 38 – 11 x 15 = 5^2

6 x 7 + 26 x 33 = 30^2
6 x 33 – 7 x 26 = 4^2

6 x 8 + 16 x 22 = 20^2
6 x 22 – 8 x 16 = 2^2

7 x 8 + 40 x 47 = 44^2
7 x 47 – 8 x 40 = 3^2

7 x 10 + 11 x 30 = 20^2
7 x 30 – 10 x 11 = 10^2

7 x 10 + 15 x 22 = 20^2
7 x 22 – 10 x 15 = 2^2

7 x 12 + 15 x 43 = 27^2
7 x 43 – 12 x 15 = 11^2

8 x 9 + 23 x 36 = 30^2
8 x 36 – 9 x 23 = 9^2

8 x 10 + 22 x 32 = 28^2
8 x 32 – 10 x 22 = 6^2

8 x 11 + 13 x 21 = 19^2
8 x 21 – 11 x 13 = 5^2

8 x 11 + 13 x 24 = 20^2
8 x 24 – 11 x 13 = 7^2

8 x 12 + 15 x 23 = 21^2
8 x 23 – 12 x 15 = 2^2

9 x 10 + 18 x 45 = 30^2
9 x 45 – 10 x 18 = 15^2

9 x 12 + 24 x 33 = 30^2
9 x 33 – 12 x 24 = 3^2

10 x 13 + 22 x 35 = 30^2
10 x 35 – 13 x 22 = 8^2

10 x 13 + 30 x 49 = 40^2
10 x 49 – 13 x 30 = 10^2

10 x 14 + 20 x 38 = 30^2
10 x 38 – 14 x 20 = 10^2

11 x 12 + 24 x 32 = 30^2
11 x 32 – 12 x 24 = 8^2

11 x 14 + 21 x 30 = 28^2
11 x 30 – 14 x 21 = 6^2

11 x 14 + 35 x 46 = 42^2
11 x 46 – 14 x 35 = 4^2

11 x 15 + 21 x 44 = 33^2
11 x 44 – 15 x 21 = 13^2

11 x 16 + 19 x 35 = 29^2
11 x 35 – 16 x 19 = 9^2

12 x 13 + 23 x 35 = 31^2
12 x 35 – 13 x 23 = 11^2

12 x 15 + 33 x 48 = 42^2
12 x 48 – 15 x 33 = 9^2

12 x 16 + 23 x 39 = 33^2
12 x 39 – 16 x 23 = 10^2

12 x 16 + 32 x 44 = 40^2
12 x 44 – 16 x 32 = 4^2

14 x 20 + 30 x 44 = 40^2
14 x 44 – 20 x 30 = 4^2

16 x 19 + 20 x 24 = 28^2
16 x 24 – 19 x 20 = 2^2

16 x 19 + 20 x 36 = 32^2
16 x 36 – 19 x 20 = 14^2

16 x 22 + 26 x 42 = 38^2
16 x 42 – 22 x 26 = 10^2

16 x 22 + 26 x 48 = 40^2
16 x 48 – 22 x 26 = 14^2

16 x 24 + 30 x 46 = 42^2
16 x 46 – 24 x 30 = 4^2

18 x 20 + 31 x 40 = 40^2
18 x 40 – 20 x 31 = 10^2

20 x 27 + 33 x 45 = 45^2
20 x 45 – 27 x 33 = 3^2

26 x 29 + 39 x 50 = 52^2
26 x 50 – 29 x 39 = 13^2

27 x 28 + 35 x 47 = 49^2
27 x 47 – 28 x 35 = 17^2

27 x 28 + 45 x 48 = 54^2
27 x 48 – 28 x 45 = 6^2

28 x 31 + 32 x 39 = 46^2
28 x 39 – 31 x 32 = 10^2

29 x 30 + 33 x 35 = 45^2
29 x 35 – 30 x 33 = 5^2

32 x 38 + 40 x 48 = 56^2
32 x 48 – 38 x 40 = 4^2

Paul.

• benvitalis says:

(a^2 + c^2)(b^2 + d^2) = (a*b + c*d)^2 + (a*d – b*c)^2

a*b + c*d, a*d – b*c
4 x 5 + 11 x 16 = 14^2
4 x 16 – 5 x 11 = 3^2
(4^2 + 11^2)(5^2 + 16^2) = 14^4 + 3^4
sum of two 4-th powers

• benvitalis says:

Extension to that: can you find a,b,c,d
(a^2 + c^2)(b^2 + d^2) is a square
(a^2 + c^2)(b^2 + d^2) = N^2 so we would have
N^2 = x^4 + y^4
a*b + c*d = x^2, a*d – b*c = y^2