Last edited by JaneFairfax (2009-03-23 02:33:42)

I misunderstood the term “square free”. I thought it was synonymous with “non perfect square” (and therefore excluded 1).

I have rephrased the question. It shouldn’t be ambiguous now.

why has no one tried #1 ???

newayz ...

we can rewrite the equn as ... (n-2)(n-1)(n²-2)(n+1)(n+2)

we can clearly see that this expression will be div for n= 7k+1,7k+2,7k+5,7k+6 ... as in each of these casesone of n-1,n-2,n+1,n+2 ... contributes a factor of 7

for 7k+3 and 7k+4 we can find, by substitution, that n²-2 donates a factor of 7.

hence the expression is div by 7 ...

now, if n-1 and n+1 are even and as they are consecutive nos. we can write them as ... 2a,2a+2
now, 2a*(2a+2) = 4a*(a+1)
now a(a+1) will be div by 2! hence div of 8 is also true when n-1 is even

now if n-2 is even ... so is n²-2 and n+2 ....
3 even nos hence three multiples of 8 ... hence div by 8 ....



thus the exp is always div by 56 ...

Last edited by smiyc86 (2009-04-01 17:32:37)

@9 ... done