1I need more insight than just a random example.
Find out integers such that (a+b+c+d)2=a3+b3+c3+d3
Let me make it simpler ..
find integers so that
(a1+a2+a3...an)2=a13+a23+a33...an3
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9 Answers
341From my readings: the unique solution set (up to permutations) for n integers is {1,2,3,...,n}
Also, your wording invited an answer like Jai's. Something on the lines of, "for any n find integers a1,... such that" would have prompted a search for a general solution as you intended.
1Exactly Prophet Sir
Can any one tell me why is it always the natural numbers ?
1Because summation of cubes of first " n " natural numbers is exactly ( and interestingly ) the square of the sum of first " n ' natural numbers .
1In other words ,
1 3 + 2 3 + 3 3 + ...........+ n 3 = [ n ( n + 1 )2 ] 2
1 + 2 + 3 + 4 + 5 + ............+ n = n ( n + 1 )2
341He meant to ask, can you prove that for any n, its is only the numbers 1,2,...,n that have this property and not any other collection of n integers.
262a collection of any n integers other than consecutive natural nos. wont hav this property....
a3+(a+1)3+(a+2)3+(a+3)3 ={ (a+3)(a+4)2} 2 - {a(a+1)2}2
a+ (a+1) + (a+2) + (a+3) = (a+3)(a+4)2 - a(a+1)2
hence the given condition will never hold unless we take the natural nos. from 1
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