# find four positive integers a,b,c,d whose 5th powers sum to the 5th power of another integer e, i.e. find five integers a,b,c,d,e such that a^5+b^5+c^5+d^5=e^5. Anonymous

I am working on Sage Math online , on Windows 10

In 1769, Lenhard Euler conjectured that at least n nth powers are required to obtain a sum that is itself an nth power for n>2. Disprove Euler's conjecture by writing an appropriate function and using it to find four positive integers a,b,c,d whose 5th powers sum to the 5th power of another integer e, i.e. find five integers a,b,c,d,e such that a^5+b^5+c^5+d^5=e^5.

I have tried this, but it seems like not given me what I wanted it. By the way, I am not allow to use "IF" Statement at all.

def calculateValues():

sumoffourvalues = 0

for i in range(2,6):

#display each valueS

print (i)

#calculate each value its power

eachpower = pow(i,5)

#sum of power values of four elements

sumoffourvalues = sumoffourvalues + eachpower

print ("power of sumofforvalues",sumoffourvalues)

#calculating power of fifth element

fifthelementvalue = pow(6,5)

print ("power of fifthelementvalue ",fifthelementvalue)

#Consider like below this way

#a5+b5+c5+d5−e5=0

resultantvalue = sumoffourvalues -fifthelementvalue

print("resultant value is ",resultantvalue)

calculateValues()

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The following code finds the two solutions in the range of natural numbers $<11\cdot 31$ in a relatively quick manner.

def PossibilitiesMod( p, ordered=True ):
"""Compute all relevant (a,b,c,d) mod p
"""
F = GF( p )
powers = list( set( [ a^5 for a in F ] ) )

possibilities = [ (a, b, c, d)
for a in F for b in F for c in F for d in F
if a^5 + b^5 + c^5 + d^5 in powers ]
if ordered:
possibilities = [ (a, b, c, d)
for (a, b, c, d) in possibilities
if a.lift() <= b.lift() <= c.lift() <= d.lift() ]
possibilities.sort()
return possibilities

p1 = 11      ; p2 = 31      ;
F1 = GF( p1 ); F2 = GF( p2 );

matchList1 = PossibilitiesMod( p1, ordered=False )
matchList2 = PossibilitiesMod( p2, ordered=True  )

lift_dic = {}
for x in [ 1 .. p1*p2 ]:
lift_dic[ ( F1(x), F2(x) ) ] = x

fifthPowers = [ a^5 for a in range( 1, ( p1 * p2 *5^(1/5) ) .n() ) ]
solutions   = []

count = 0
for a1, b1, c1, d1 in matchList1:
count += 1
if count % 1000 == 0:    print count
for a2, b2, c2, d2 in matchList2:

a = lift_dic[ ( a1, a2 ) ]
b = lift_dic[ ( b1, b2 ) ]
c = lift_dic[ ( c1, c2 ) ]
d = lift_dic[ ( d1, d2 ) ]

if a^5 + b^5 + c^5 + d^5 in fifthPowers:
print "\n\n\n!!! SOLUTION !!! ( %s , %s , %s , %s )\n\n\n" % ( a, b, c, d )
solutions.append( (a,b,c,d) )


Namely:

( 220 , 168 , 266 , 54 )
( 133 , 110 , 84 , 27 )


The last one is of course:

Lander and Parkin, 1966, "a direct search on the CDC 6600..."

(The first solution above is "twice the second one". Sorry.) To have a feasible search we have restricted first to relevant matches modulo some prime numbers, in our case $11$ and $31$. (Chosen so that the Euler indicator function on them delivers multiples of $5$, the power appearing in the equation.)

For the other related problems with higher powers this is of course too naive and slow...

https://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture

http://www.cs.man.ac.uk/~rizos/EqualSums/equalsums.html

http://euler.free.fr/index.htm

(I'm afraid, this answers an other question, not explicitly the one posted above. The "no-if-restriction" is in m.h.o. a good joke, i've got the same feeling as facing "show this inequality without using algebra" some decades ago.)

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