Question

Prove that the product of 4 consecutive natural numbers cannot be a perfect cube.

Answer 1

n+n+1+n+2+n+3 = 4n + 3

a perfect cube (n+a)^3 = n^3 +3n^2a + 3na^2 + a^3

compare the two rewults

Answer 2

oh, product ....

Answer 3

you get n^4 + ....

Answer 4

ok

Answer 5

n * n+1 * n+2 * n+3 = n^4 + ... for n >1

Answer 6

\[n^4+6n^3+11n^2+6n\]

Answer 7

Ok Thank you sir

Answer 8

we could possibly compare this with x^3 and see if they are equal at some n in N

Answer 9

ok i'll do that

Answer 10

\[y=n^4+6n^3+11n^2+6n~:~y=n^3\]
\[n^4+6n^3+11n^2+6n=n^3\]
\[n^4+5n^3+11n^2+6n=0\]
\[n(n^3+5n^2+11n+6)=0\]
\[n^3+5n^2+11n+6\]
by descartes sign rule, this has no positive solutons

Answer 11

ok

Answer 12

http://openstudy.com/study#/updates/518c0588e4b062a8d1d98c12