All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Prime as sum of Cubes of terms of AP? (Posted on 2008-03-02)
Let Si be the ith term of an Arithmetic Progression whose 1st term is a and common difference d. Show that for any 2 positive integers m,n(>m), Σ(i:m to n)(Si)3 can't be a prime number.

Note: a,d are positive integers.

 See The Solution Submitted by Praneeth Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 hint- no spoiler | Comment 3 of 4 |

THE SUM OF CUBES (   Si)^3 of AP is always divisible by the sum of correspondng members  of  that AP  - therefore it cannot be prime.

E.G.

given  2,5,8,11,14 etc

2^3 +5^3    is divisible by  2+5

2^3 +5^3 + 8^3   is divisible by  2+5+8

2^3 +5^3 + 8^3+ 11^3+14^3   is divisible by  2+5+8+11+14

ETC

Now- go and prove it formally

Edited on March 7, 2008, 7:43 am
 Posted by Ady TZIDON on 2008-03-07 07:41:28

 Search: Search body:
Forums (0)