Number Theory Problem No.3
We know that this Diophantine equation (eq.(1)) :
a n + b n = c n (1)
( n are positive integer)
For n = 2 there are infinitely many nonzero natural solutions (a, b, c): the Pythagorean triples . For n > 2 , Fermat's Last Theorem (initially claimed in 1637 by Fermat and proved by Andrew Wiles in 1995) states there are no positive integer solutions ( a , b , c ) .
But does this equation (eq.(2)) have nonzero natural solution?
a 3 + b 3 + c 3 = d 3 (2)
Bonus: Does the equation a 1 n + a 2 n + . . . + a m n = b n have positive integer solutions ( a 1 , a 2 , . . . a m , b ) for m , n are natural number greater than 2? Explain in the discussion.
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4 solutions
This is not an explanation but my question is can you make b n from a 1 n + a 2 n + a 3 n + … + a n n
Rule:
a 1 , a 2 , a 3 , … , a n , b and n have to be positive integers
Tell me a proof if you can
3 3 + 4 3 + 5 3 = 6 3
I found 1 counterexample ( 3 , 4 , 5 , 6 ) for n = 3 .
Further explanation is for you.
Another is ( 1 , 6 , 8 , 9 ) . There's a lot more information here , including a couple of formulas to generate these quadruples.
( 3 d ) 3 + ( 4 d ) 3 + ( 5 d ) 3 = ( 6 d ) 3
I think the solution for any k ≥ 3 : n = 1 ∑ k ( 2 + n ) k = ( 3 + k ) k