## System {a^2 + b^2 + c^2 = 2*x^2, a^4 + b^4 + c^4 = 2*x^4}

Give integral solutions of the system:

$a^2 \; + \; b^2 \; + \; c^2 \; = \; 2 \, x^2$
$a^4 \; + \; b^4 \; + \; c^4 \; = \; 2 \, x^4$

Note that

$1/2 \; (a^2+b^2+(a+b)^2) \; = \; a^2+a b+b^2$
$1/2 \; (a^4+b^4+(a+b)^4) \; = \; (a^2+a b+b^2)^2$

$(1/2 \; ( \,a^2+b^2+(b-a)^2 \,) \,)^2 \; = \; 1/2 \; ( \,a^4+b^4+(b-a)^4 \,)$
$(1/2 \; ( \,a^2+b^2+(a-b)^2 \,))^2 \; = \; 1/2 \; ( \,a^4+b^4+(a-b)^4 \,)$
$(1/2 \; ( \,a^2+b^2+(a+b)^2 \,)^2 \; = \; 1/2 \; ( \,a^4+b^4+(a+b)^4 \,)$

math grad - Interest: Number theory
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### 2 Responses to System {a^2 + b^2 + c^2 = 2*x^2, a^4 + b^4 + c^4 = 2*x^4}

1. Paul says:

Here are those with a < b < c <= 100

3^2 + 5^2 + 8^2 = 2 x 7^2
3^4 + 5^4 + 8^4 = 2 x 7^4

5^2 + 16^2 + 21^2 = 2 x 19^2
5^4 + 16^4 + 21^4 = 2 x 19^4

6^2 + 10^2 + 16^2 = 2 x 14^2
6^4 + 10^4 + 16^4 = 2 x 14^4

7^2 + 8^2 + 15^2 = 2 x 13^2
7^4 + 8^4 + 15^4 = 2 x 13^4

7^2 + 33^2 + 40^2 = 2 x 37^2
7^4 + 33^4 + 40^4 = 2 x 37^4

9^2 + 15^2 + 24^2 = 2 x 21^2
9^4 + 15^4 + 24^4 = 2 x 21^4

9^2 + 56^2 + 65^2 = 2 x 61^2
9^4 + 56^4 + 65^4 = 2 x 61^4

10^2 + 32^2 + 42^2 = 2 x 38^2
10^4 + 32^4 + 42^4 = 2 x 38^4

11^2 + 24^2 + 35^2 = 2 x 31^2
11^4 + 24^4 + 35^4 = 2 x 31^4

11^2 + 85^2 + 96^2 = 2 x 91^2
11^4 + 85^4 + 96^4 = 2 x 91^4

12^2 + 20^2 + 32^2 = 2 x 28^2
12^4 + 20^4 + 32^4 = 2 x 28^4

13^2 + 35^2 + 48^2 = 2 x 43^2
13^4 + 35^4 + 48^4 = 2 x 43^4

14^2 + 16^2 + 30^2 = 2 x 26^2
14^4 + 16^4 + 30^4 = 2 x 26^4

14^2 + 66^2 + 80^2 = 2 x 74^2
14^4 + 66^4 + 80^4 = 2 x 74^4

15^2 + 25^2 + 40^2 = 2 x 35^2
15^4 + 25^4 + 40^4 = 2 x 35^4

15^2 + 48^2 + 63^2 = 2 x 57^2
15^4 + 48^4 + 63^4 = 2 x 57^4

16^2 + 39^2 + 55^2 = 2 x 49^2
16^4 + 39^4 + 55^4 = 2 x 49^4

17^2 + 63^2 + 80^2 = 2 x 73^2
17^4 + 63^4 + 80^4 = 2 x 73^4

18^2 + 30^2 + 48^2 = 2 x 42^2
18^4 + 30^4 + 48^4 = 2 x 42^4

19^2 + 80^2 + 99^2 = 2 x 91^2
19^4 + 80^4 + 99^4 = 2 x 91^4

20^2 + 64^2 + 84^2 = 2 x 76^2
20^4 + 64^4 + 84^4 = 2 x 76^4

21^2 + 24^2 + 45^2 = 2 x 39^2
21^4 + 24^4 + 45^4 = 2 x 39^4

21^2 + 35^2 + 56^2 = 2 x 49^2
21^4 + 35^4 + 56^4 = 2 x 49^4

22^2 + 48^2 + 70^2 = 2 x 62^2
22^4 + 48^4 + 70^4 = 2 x 62^4

24^2 + 40^2 + 64^2 = 2 x 56^2
24^4 + 40^4 + 64^4 = 2 x 56^4

26^2 + 70^2 + 96^2 = 2 x 86^2
26^4 + 70^4 + 96^4 = 2 x 86^4

27^2 + 45^2 + 72^2 = 2 x 63^2
27^4 + 45^4 + 72^4 = 2 x 63^4

28^2 + 32^2 + 60^2 = 2 x 52^2
28^4 + 32^4 + 60^4 = 2 x 52^4

30^2 + 50^2 + 80^2 = 2 x 70^2
30^4 + 50^4 + 80^4 = 2 x 70^4

32^2 + 45^2 + 77^2 = 2 x 67^2
32^4 + 45^4 + 77^4 = 2 x 67^4

33^2 + 55^2 + 88^2 = 2 x 77^2
33^4 + 55^4 + 88^4 = 2 x 77^4

35^2 + 40^2 + 75^2 = 2 x 65^2
35^4 + 40^4 + 75^4 = 2 x 65^4

36^2 + 60^2 + 96^2 = 2 x 84^2
36^4 + 60^4 + 96^4 = 2 x 84^4

40^2 + 51^2 + 91^2 = 2 x 79^2
40^4 + 51^4 + 91^4 = 2 x 79^4

42^2 + 48^2 + 90^2 = 2 x 78^2
42^4 + 48^4 + 90^4 = 2 x 78^4

Paul.

• benvitalis says:

Note that
$(1/2 \; ( \,a^2+b^2+(b-a)^2 \,) \,)^2 \; = \; 1/2 \; ( \,a^4+b^4+(b-a)^4 \,)$
$(1/2 \; ( \,a^2+b^2+(a-b)^2 \,))^2 \; = \; 1/2 \; ( \,a^4+b^4+(a-b)^4 \,)$
$(1/2 \; ( \,a^2+b^2+(a+b)^2 \,)^2 \; = \; 1/2 \; ( \,a^4+b^4+(a+b)^4 \,)$