Cube Hunting
Find all solutions to the equation 2 n 3 + 6 n 2 + 1 0 2 n + 9 8 = m 3 m , n ∈ Z . If the answers are ( m 1 , n 1 ) , ( m 2 , n 2 ) , … , ( m N , n N ) , evaluate S = j = 1 ∑ N ( m j 2 + n j 2 ) .
The answer is 163.
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3 solutions
The equation may be written as (n+1)^3+48(n+1)=m3 which reduces to t^3+3t=4s^3 where 4t=(n+1)following successive substitutions . Now the equation t^3+3t=4s^3 seems to be only possible for t=-1,0,1( I do not have a proof of this) but working back up gives the correct answer
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Having got to t 3 + 3 t , think ( t + 1 ) 3 + ( t − 1 ) 3 ...
Another thought: Would it be possible to prove that t^3+3t=4s^3 has no other solution by employing the method of infinite descent noting that we already have the minimal solution in integers
did the same way but, there is no need to check so many cases, three cases are enough;
case 1: when m=0
case 2: when n+5=0
case 3: when n-3=0
the above are the only solutions by Fermat's last theorem.
I do not have much understanding of Theory of Numbers but using elementary methods:- L e t , f ( n ) = 2 n 3 + 6 n 2 + 1 0 2 n + 9 8 . f ( − 1 ) = − 2 + 6 − 1 0 2 + 9 8 = 0 , ∴ n = − 1 , a n d m = 0 , i m p l i e s ( m , n ) = ( 0 , − 1 ) i s a s o l u t i o n . m m u s t b e a n i n t e g e r . S o , 3 2 n 3 + 6 n 2 + 1 0 2 n + 9 8 t o o m u s t b e a n i n t e g e r . T r i e d f o r n = − 5 , − 4 , − 3 , − 2 , 1 , 2 , 3 a n d g o t t w o m o r e s o l u t i o n s , ( m , n ) i s ( − 8 , − 5 ) , ( 0 , − 1 ) , ( 8 , 3 ) s o s t o p p e d . S i n c e i t i s a c u b i c , w e g o t a l l t h r e e s o l u t i o n o f f ( n ) − m 3 . A n s w e r = 0 2 + ( − 1 ) 2 + 8 2 + 3 2 + ( − 8 ) 2 + ( − 5 ) 2 = 1 6 3 .
The existence of only three integer solutions to the equation f ( n ) = m 3 has nothing to do with the fact that it is cubic. As a trivial counterexample, consider the equation n 3 = m 3 . Although it is cubic, there are infinitely many integer solutions to it.
Since 2 n 3 + 6 n 2 + 1 0 2 n + 9 8 = ( n + 5 ) 3 + ( n − 3 ) 3 we are trying to solve the Diophantine equation ( n + 5 ) 3 + ( n − 3 ) 3 = m 3 m , n ∈ Z ( ⋆ ) It is well-known that the equation x 3 + y 3 = z 3 has no solutions in positive integers x , y , z (or in negative integers), and hence ( ⋆ ) has no solutions with n ≥ 4 or n ≤ − 6 .
Checking the cases n = − 5 , − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 gives us that the only possible solutions are ( m , n ) = ( 8 , 3 ) , ( 0 , − 1 ) , ( − 8 , − 5 ) . Thus the required answer is S = 8 2 + 3 2 + 0 2 + ( − 1 ) 2 + ( − 8 ) 2 + ( − 5 ) 2 = 1 6 3