Sharedsupport / 2017-03-14-101118-sums-of-fourth-powers.sagewsOpen in CoCalc
Authors: Harald Schilly, ℏal Snyder, William A. Stein
License: GNU General Public License v3.0
Description: Examples for support purposes.

Question: I am looking for two integers, aa and bb, such that a4+b4=11572060353961555386606814001a^4+ b^4 = 11572060353961555386606814001. Can anyone help me to solve this problem using SageMath .

n = 11572060353961555386606814001 factor(n)
101 * 44621 * 10147901 * 253031037087781
# n is a sum of two *squares*: sum_of_k_squares(2, n)
(70463059530200, 81283562887001)
#but not a perfect square N(sqrt(70463059530200))
8.39422775067486e6
for p, _ in factor(n): print p, p%4
101 1 44621 1 10147901 1 253031037087781 1

Because each pp dividing nn exactly divides nn and is 11 mod 44, there are 16=242^4 ways to write nn as a sum of two squares. If you write nn as a sum of two fourth powers, that's just one of these: (a2)2+(b2)2=n(a^2)^2 + (b^2)^2 = n

K.<i> = QuadraticField(-1) v = [factor(K(p)) for p, _ in factor(n)] v
[(-1) * (i - 10) * (i + 10), (-10*i + 211) * (10*i + 211), (-1099*i + 2990) * (1099*i + 2990), (-5563791*i + 14902190) * (5563791*i + 14902190)]
v[0][0][0]
i - 10
for c in cartesian_product_iterator([[0,1]]*4): rep = prod(v[i][k][0] for i, k in enumerate(c)) print c, rep if is_square(rep[0]): print "FOUND ONE! ", rep
(0, 0, 0, 0) 81283562887001*i - 70463059530200 (0, 0, 0, 1) 15216760633199*i - 106491833253980 (0, 0, 1, 0) 16301319409199*i - 106331215263820 (0, 0, 1, 1) -57372053092999*i - 90997295992000 (0, 1, 0, 0) 74255239366801*i - 77834566746020 (0, 1, 0, 1) 5077165185799*i - 107453630686160 (0, 1, 1, 0) 6172053092999*i - 107396304008000 (0, 1, 1, 1) -65720906289199*i - 85163506447820 (1, 0, 0, 0) -65720906289199*i + 85163506447820 (1, 0, 0, 1) 6172053092999*i + 107396304008000 (1, 0, 1, 0) 5077165185799*i + 107453630686160 (1, 0, 1, 1) 74255239366801*i + 77834566746020 (1, 1, 0, 0) -57372053092999*i + 90997295992000 (1, 1, 0, 1) 16301319409199*i + 106331215263820 (1, 1, 1, 0) 15216760633199*i + 106491833253980 (1, 1, 1, 1) 81283562887001*i + 70463059530200
# e.g., 5077165185799^2 + 107453630686160^2
11572060353961555386606814001

As you can see, we didn't find any with the rep as a sum of two squares with each a square itself. So there are no such a,ba,b.

Same problem by brute force search...

%python from math import sqrt def sum_of_two_fourth_powers(m): b = int(sqrt(sqrt(m))) print(b) for i in xrange(b): s = sqrt(sqrt(m-i**4)) if int(s) == s: return i, int(s)
# test the above function %time sum_of_two_fourth_powers(848^4 + 798^4)
980 (798, 848) CPU time: 0.00 s, Wall time: 0.00 s
# now try it on yours... and get nothing. %time sum_of_two_fourth_powers(11572060353961555386606814001)
10371765 CPU time: 23.28 s, Wall time: 23.68 s