Prove that the product of four consecutive natural numbers, plus 1, is a perfect square?

Answer 1

Let x be the smallest of four consecutive numbers, which would then be x; x + 1; x + 2; x + 3: Then their product, plus 1, is

= x(x + 3)(x + 2)(x + 1) + 1

= (x^2 + 3x)(x^2 + 3x + 2) + 1

= ((x^2 + 3x + 1) - 1)((x^2 + 3x + 1) + 1) + 1

= ((x^2 + 3x + 1)^2 - 1^2)+1 = (x^2 + 3x + 1)^2;

which is a perfect square.

Answer 2

i do no longer comprehend the respond off the suitable of my head, yet i comprehend how i might attack it. a great integer is a suitable cube if and on condition that: once you ingredient it uniquely into primes, each and every of the exponents is divisible by 3. So look on the series of four consecutive integers. 2 would be wonderful, one would be even yet no longer divisible by 2^2, and one would be divisible by 2^x for some x>one million. So x+one million could be divisible by 3. Skipping over 3, the two between the integers is divisible by 5 or none is. nicely, if there is one divisible by 5, its factorization could comprise a means of 5 it quite is divisible by 3 ... Ya comprehend, it extremely is not getting everywhere. the subsequent attack that involves concepts is to write down it as, say (x-one million)x(x+one million)(x+2) = say (x^3 - x)(x + 2), and seek for concept. Hmm. i'm stumped on the 2d. ----------------------------------- EDIT: in the respond under, there seems to have been an implied "with out loss of generality, the 1st of the 4 words is wonderful." i assume it quite is okay, because of the fact the evidence isn't tormented by no remember if the numbers are valuable or damaging, even in spite of the shown fact that it nevertheless regarded somewhat rapid. -- Oh, that replaced right into a reasonable typo. What extremely replaced into meant, i think of, replaced right into a concentration on the two wonderful words. the reason they have been chosen to be wonderful is that, whilst the powers of three are got rid of, what's left over could be suitable cubes. anyhow, I think of a extra ordinary evidence replaced into needed ...

Answer 3

Start with:
n(n+1)(n+2)(n+3)+1

Expand it out. Then factorise it into the form K².
QED.

ie.,

n(n+1)(n+2)(n+3)+1
=
n^4 + 6n^3 + 11n^2 + 6n + 1
=
(n²+3n+1)²

Thus it's a perfect square.


Doctor Q, your method is not sufficient proof. If you implemented a general term, then you could do it by induction.

Answer 4

1 x 2 x 3 x 4 = 24, plus 1 = 25 (i.e. 5^2)

2 x 3 x 4 x 5 = 120, plus 1 = 121 (i.e. 11^2)

3 x 4 x 5 x 6 = 360, plus 1 = 361 (i.e. 19^2)

and so on.....

Answer 5

Don't listen to Doctor Q, to prove it that way, you'd have to do that for every combination of numbers, of which there are infinity.
Ozo is right, I've just varified it myself, it's about 12 lines of algebra.

edit: or I could get ninja'd while I'm doing my algebra.

Answer 6

(1 x 2 x 3 x 4) + 1 = 25

5^2 = 25

similarly

n(n+1)(n+2)(n+3) = .........multiply this out

And rearrange it as a square.

Answer 7

X*(X+1)*(X+2)*(X+3)+1

(X^2+X)*(X^2+5X+6)+1

X^4+5X^3+6X^2+X^3+5X^2+6X+1

X^4+6X^3+11X^2+6X+1

(X^2+3X+1)^2