**(1) a^4 + 2 b^4 + 2 c^4 = d^4**

**(2) a^5 + 2 b^5 + 2 c^5 = d^5**

**(1)**

**2^4 + 2^4 + 3^4 + 4^4 + 4^4 = 625 = 5^4**

In other words,

**[2, 2, 3, 4, 4] = [5]**

Note that 3^2 + 4^2 = 5^2, (3, 4, 5) is a primitive Pythagorean triple

Find other Pythagorean-like solutions.

that is to say, find (a, b, c, d) so that

a^4 + 2 b^4 + 2 c^4 = d^4

and

b^2 + c^2 = d^2 or a^2 + b^2 = d^2

**Pythagorean triple**

http://en.wikipedia.org/wiki/Pythagorean_triple

**[4, 4, 6, 8, 8] = [10]**

2*(4^4) + 6^4 + 2*(8^4) = 10^4

[6, 8, 10] is a multiple of [3, 4, 5]

**[6, 6, 9, 12, 12] = [15]**

2*(6^4) + 9^4 + 2*(12^4) = 15^4

[9, 12, 15] is a multiple of [3, 4, 5]

**[8, 8, 12, 16, 16] = [20]**

2*(8^4) + 12^4 + 2*(16^4) = 20^4

[12, 16, 20] is a multiple of [3, 4, 5]

**[10, 10, 15, 20, 20] = [25]**

2*(10^4) + 15^4 + 2*(20^4) = 25^4

[15, 20, 25] is a multiple of [3, 4, 5]

**[12, 12, 18, 24, 24] = [30]**

2*(12^4) + 18^4 + 2*(24^4) = 30^4

[18, 24, 30] is a multiple of [3, 4, 5]

**[14, 14, 21, 28, 28] = [35]**

2*(14^4) + 21^4 + 2*(28^4) = 35^4

[21, 28, 35] is a multiple of [3, 4, 5]

**[16, 16, 24, 32, 32] = [40]**

2*(16^4) + 24^4 + 2*(32^4) = 40^4

[24, 32, 40] is a multiple of [3, 4, 5]

**[18, 18, 27, 36, 36] = [45]**

2*(18^4) + 27^4 + 2*(36^4) = 45^4

[27, 36, 45] is a multiple of [3, 4, 5]

**and so on.**

Can you find another **primitive** solution, that is, a solution not a multiple of [3, 4, 5] ?

**(2)**

**526^5 + 526^5 + 1349^5 + 1349^5 + 1355^5 = 1685^5**

( = 13583119642053125)

Can you find other solutions?

——————————————

There are 16 primitive Pythagorean triples with c ≤ 100 :

(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)

(20, 21, 29) (12, 35, 37) (9, 40, 41) (28, 45, 53)

(11, 60, 61) (16, 63, 65) (33, 56, 65) (48, 55, 73)

(13, 84, 85) (36, 77, 85) (39, 80, 89) (65, 72, 97)

[2, 2, 2, 2, 2, 2, 2, 2, 8, 8, 8, 8, 8, 8, 8, 8, 15] = [17]

8*(2^4) + 8*(8^4) + 15^4 = 17^4

#!/usr/bin/python3

# David

# a^k + 2b^k + 2c^k = d^k, where k = 4, 5

# (1) a^4 + 2 b^4 + 2 c^4 = d^4

# (2) a^5 + 2 b^5 + 2 c^5 = d^5

import math

MAX = 2

for k in range(4,6):

for a in range(1,10**MAX):

for b in range(1,10**MAX):

for c in range(1,10**MAX):

for d in range(1,10**MAX):

if (

(a**k)+(2*(b**k))+(2*(c**k)) == d**k and

(math.hypot(a,b) or math.hypot(b,c)) == d

):

print(‘Is ({0},{1},{3}) a Pythagorean-like triple? => {0}^{4} + 2*({1}^{4}) + 2*({2}^{4}) = {3}^{4}’.format(a,b,c,d,k))

# some answers re (1) ….

Is (3,4,5) a Pythagorean-like triple? => 3^4 + 2*(4^4) + 2*(2^4) = 5^4

Is (6,8,10) a Pythagorean-like triple? => 6^4 + 2*(8^4) + 2*(4^4) = 10^4

Is (9,12,15) a Pythagorean-like triple? => 9^4 + 2*(12^4) + 2*(6^4) = 15^4

Is (12,16,20) a Pythagorean-like triple? => 12^4 + 2*(16^4) + 2*(8^4) = 20^4

Is (15,20,25) a Pythagorean-like triple? => 15^4 + 2*(20^4) + 2*(10^4) = 25^4

Is (18,24,30) a Pythagorean-like triple? => 18^4 + 2*(24^4) + 2*(12^4) = 30^4

Is (21,28,35) a Pythagorean-like triple? => 21^4 + 2*(28^4) + 2*(14^4) = 35^4

Is (24,32,40) a Pythagorean-like triple? => 24^4 + 2*(32^4) + 2*(16^4) = 40^4

Is (27,36,45) a Pythagorean-like triple? => 27^4 + 2*(36^4) + 2*(18^4) = 45^4

Is (30,40,50) a Pythagorean-like triple? => 30^4 + 2*(40^4) + 2*(20^4) = 50^4

Is (33,44,55) a Pythagorean-like triple? => 33^4 + 2*(44^4) + 2*(22^4) = 55^4

Is (36,48,60) a Pythagorean-like triple? => 36^4 + 2*(48^4) + 2*(24^4) = 60^4

Is (39,52,65) a Pythagorean-like triple? => 39^4 + 2*(52^4) + 2*(26^4) = 65^4

Is (42,56,70) a Pythagorean-like triple? => 42^4 + 2*(56^4) + 2*(28^4) = 70^4

Is (45,60,75) a Pythagorean-like triple? => 45^4 + 2*(60^4) + 2*(30^4) = 75^4

Is (48,64,80) a Pythagorean-like triple? => 48^4 + 2*(64^4) + 2*(32^4) = 80^4

Is (51,68,85) a Pythagorean-like triple? => 51^4 + 2*(68^4) + 2*(34^4) = 85^4

Is (54,72,90) a Pythagorean-like triple? => 54^4 + 2*(72^4) + 2*(36^4) = 90^4

Is (57,76,95) a Pythagorean-like triple? => 57^4 + 2*(76^4) + 2*(38^4) = 95^