## Make {(a^2 + 12*b^2), (12*a^2 + b^2)} squares

Find distinct positive integers   $a, \; b, \; c, \; d$   such that

$a^2 \; + \; 12 \, b^2 \; = \; c^2$
$12 \, a^2 \; + \; b^2 \; = \; d^2$

Paul found

$\{a, \; b, \; c, \; d \} \; = \; \{1, \; 2, \; 7, \; 4 \}$
$\{a, \; b, \; c, \; d \} \; = \; \{2, \; 1, \; 4, \; 7 \}$

and their multiples

$\{a, \; b, \; c, \; d \} = \; \{2, \; 4, \; 14, \; 8 \}$
$\{a, \; b, \; c, \; d \} = \; \{3, \; 6, \; 21, \; 12 \}$
$\{a, \; b, \; c, \; d \} = \; \{4, \; 2, \; 8, \; 14 \}$
$\{a, \; b, \; c, \; d \} = \; \{4, \; 8, \; 28, \; 16 \}$
$\{a, \; b, \; c, \; d \} = \; \{5, \; 10, \; 35, \; 20 \}$
$\{a, \; b, \; c, \; d \} = \; \{6, \; 3, \; 12, \; 21 \}$
$\{a, \; b, \; c, \; d \} = \; \{6, \; 12, \; 42, \; 24 \}$
$\{a, \; b, \; c, \; d \} = \; \{7, \; 14, \; 49, \; 28 \}$
$\{a, \; b, \; c, \; d \} = \; \{8, \; 4, \; 16, \; 28 \}$
$\{a, \; b, \; c, \; d \} = \; \{8, \; 16, \; 56, \; 32 \}$
$\{a, \; b, \; c, \; d \} = \; \{9, \; 18, \; 63, \; 36 \}$
$\{a, \; b, \; c, \; d \} = \; \{10, \; 5, \; 20, \; 35 \}$
$\{a, \; b, \; c, \; d \} = \; \{10, \; 20, \; 70, \; 40 \}$
$\{a, \; b, \; c, \; d \} = \; \{11, \; 22, \; 77, \; 44 \}$
$\{a, \; b, \; c, \; d \} = \; \{12, \; 6, \; 24, \; 42 \}$
$\{a, \; b, \; c, \; d \} = \; \{12, \; 24, \; 84, \; 48 \}$
$\{a, \; b, \; c, \; d \} = \; \{13, \; 26, \; 91, \; 52 \}$
$\{a, \; b, \; c, \; d \} = \; \{14, \; 7, \; 28, \; 49 \}$
$\{a, \; b, \; c, \; d \} = \; \{14, \; 28, \; 98, \; 56 \}$
$\{a, \; b, \; c, \; d \} = \; \{16, \; 8, \; 32, \; 56 \}$
$\{a, \; b, \; c, \; d \} = \; \{16, \; 8, \; 32, \; 56 \}$
$\{a, \; b, \; c, \; d \} = \; \{18, \; 9, \; 36, \; 63 \}$
$\{a, \; b, \; c, \; d \} = \; \{20, \; 10, \; 40, \; 70 \}$
$\{a, \; b, \; c, \; d \} = \; \{22, \; 11, \; 44, \; 77 \}$
$\{a, \; b, \; c, \; d \} = \; \{24, \; 12, \; 48, \; 84 \}$
$\{a, \; b, \; c, \; d \} = \; \{26, \; 13, \; 52, \; 91 \}$
$\{a, \; b, \; c, \; d \} = \; \{28, \; 14, \; 56, \; 98 \}$

math grad - Interest: Number theory
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### 2 Responses to Make {(a^2 + 12*b^2), (12*a^2 + b^2)} squares

1. Paul says:

Here are those with c & d <=100

1^2 + 12*2^2 = 7^2
12*1^2 + 2^2 = 4^2

2^2 + 12*1^2 = 4^2
12*2^2 + 1^2 = 7^2

2^2 + 12*4^2 = 14^2
12*2^2 + 4^2 = 8^2

3^2 + 12*6^2 = 21^2
12*3^2 + 6^2 = 12^2

4^2 + 12*2^2 = 8^2
12*4^2 + 2^2 = 14^2

4^2 + 12*8^2 = 28^2
12*4^2 + 8^2 = 16^2

5^2 + 12*10^2 = 35^2
12*5^2 + 10^2 = 20^2

6^2 + 12*3^2 = 12^2
12*6^2 + 3^2 = 21^2

6^2 + 12*12^2 = 42^2
12*6^2 + 12^2 = 24^2

7^2 + 12*14^2 = 49^2
12*7^2 + 14^2 = 28^2

8^2 + 12*4^2 = 16^2
12*8^2 + 4^2 = 28^2

8^2 + 12*16^2 = 56^2
12*8^2 + 16^2 = 32^2

9^2 + 12*18^2 = 63^2
12*9^2 + 18^2 = 36^2

10^2 + 12*5^2 = 20^2
12*10^2 + 5^2 = 35^2

10^2 + 12*20^2 = 70^2
12*10^2 + 20^2 = 40^2

11^2 + 12*22^2 = 77^2
12*11^2 + 22^2 = 44^2

12^2 + 12*6^2 = 24^2
12*12^2 + 6^2 = 42^2

12^2 + 12*24^2 = 84^2
12*12^2 + 24^2 = 48^2

13^2 + 12*26^2 = 91^2
12*13^2 + 26^2 = 52^2

14^2 + 12*7^2 = 28^2
12*14^2 + 7^2 = 49^2

14^2 + 12*28^2 = 98^2
12*14^2 + 28^2 = 56^2

16^2 + 12*8^2 = 32^2
12*16^2 + 8^2 = 56^2

18^2 + 12*9^2 = 36^2
12*18^2 + 9^2 = 63^2

20^2 + 12*10^2 = 40^2
12*20^2 + 10^2 = 70^2

22^2 + 12*11^2 = 44^2
12*22^2 + 11^2 = 77^2

24^2 + 12*12^2 = 48^2
12*24^2 + 12^2 = 84^2

26^2 + 12*13^2 = 52^2
12*26^2 + 13^2 = 91^2

28^2 + 12*14^2 = 56^2
12*28^2 + 14^2 = 98^2

Paul.

• benvitalis says:

You found
$\{a, \; b, \; c, \; d \} \; = \; \{1, \; 2, \; 7, \; 4 \}$
$\{a, \; b, \; c, \; d \} \; = \; \{2, \; 1, \; 4, \; 7 \}$

and their multiples.

Can you find other primitive solutions?