prove that the product of 4 consecutive natural numbers is not a perfect cube?

Answer 1

let 4 consecutive numbers be n,n+1,n+2,n+3
=>p=n(n+1)(n+2)(n+3)
=>p=1/5(n+4-n+1)n(n+1)(n+2)(n+3)
=>p=1/5n(n+1)(n+2)(n+3)(n+4) - 1/5(n-1)n(n+1)(n+2)(n+3)
now put n=1,2,3,4.................and add all terms

=>p=1/5(1*2*3*4*5 - 0)
+1/5(2*3*4*5*6 - 1*2*3*4*5)
+1/5(3*4*5*6*7 - 2*3*4*5*6)
..........................................................
=>p=1/5(1*2*3*4*5 - n(n+1)(n+2)(n+3)(n+4))
=>p=1/5(120 - n(n+1)(n+2)(n+3)(n+4))
=>p is not a perfect cube, hence proved.

Answer 2

nicely, you propose beneficial integers 0*a million*2*3 = 0, that's a appropriate cube of 0 EDIT No, organic initiate from 0. Im sorry. Mathematicians use N or (an N in blackboard ambitious) to communicate to the set of all organic numbers. This set is infinite yet countable via definition. To be unambiguous approximately regardless of if 0 is roofed or no longer, each so often an index "0" is extra in the former case, and a superscript "*" is extra in the latter case: N0 = { 0, a million, 2, ... } ; N* = { a million, 2, ... }. EDIT 2 stable answer, Ben. I did somethink like that yet I couldnt discover a thank you to teach that this product wasnt a appropriate cube. thank you for this information. Ana

Answer 3

If n(n+1)(n+2)(n+3)=a^3 for some natural numbers n and a, then

n^4+6n^3+11n^2+6n=a^3

n^4+6n^3+11n^2+6n-a^3=0

From the rational zeroes theorem in college algebra, you know that if a rational (and hence natural integer) solution exists, then it must be in the form
(divisor of constant term)/(divisor of leading coefficient)
So the only possibilities here would be 1, -1, a, -a, a^2, -a^2, a^3, -a^3

When you plug these in for n, you don't get zero. Therefore a solution to this quartic equation cannot be rational so in turn a natural number solution does not exist.

>>Very true, in which case we'll have to (perhaps) resort to some result on modular arithmetic about cubes.

Answer 4

The proposed proof above seems to assume a is prime, an assumption that unfortunately makes little sense.

Answer 5

1,2,3, & 4

1 X 2 X 3 X 4 = 24

-> 24 is not a perfect cube.........

Answer 6

Well, when you add 9 to 700 and substract it from 5million. Please add the resulting number to 45 and put it in a square...you retarted ******* ontar.....