PR 4 # 33 -- The last puzzle from this series

[Wacky]

Prove that the product of 4 consecutive positive integers cannot be a perfect square.

[Citrix]

one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.

Another question. Is there a value n, n>2, such that the product of n consecutive integers is a perfect square.

[axn]

Originally Posted by Citrix

one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.

What prevents the "multiple of 4" from being a multiple of 8?

[xilman]

Originally Posted by Citrix

one of the numbers is divisible by 4 and one by 2. 8 is not a perfect square.

Another question. Is there a value n, n>2, such that the product of n consecutive integers is a perfect square.

Yes, and n can be arbitrarily large. Any range which includes zero satisfies the problem as stated.

Paul

[Citrix]

Axn1, you are right, I did not think of that.

Xilman, if you do not include zero, what t!/(t-n)! can ever be a prefect square?

[axn]

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

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

So, we need two perfect squares with a difference of 1. 0 and 1, anyone?

[alpertron]

Originally Posted by Wacky

Prove that the product of 4 consecutive positive integers cannot be a perfect square.

There are two cases: 1*2*3*4 = 24 (which is not square) and the rest:

Since the difference between any member of this set is not greater than 3, the gcd of two of these numbers cannot be greater than 3.

This means that any prime factor greater than 3, can appear only in one number of the set. It is possible that this prime factor can appear twice (or even number of times) in a particular number, but since there is at most one square number inside the set, there must be a prime > 3 that appear an odd number of times in a member of the set and does not appear in the other members of the set.

This implies that the product cannot be a perfect square.

[wblipp]

alpertron:

I don't see why your approach would prevent
the four consecutive numbers from being
27*a[sup]2[/sup]
8*b[sup]2[/sup]
c[sup]2[/sup]
6*d[sup]2[/sup]

[alpertron]

Originally Posted by wblipp

alpertron:
I don't see why your approach would prevent
the four consecutive numbers from being
27*a²
8*b²
c²
6*d²

You are right, it does not prevent it, but your sequence cannot exist:

27 a²+1 = 8 b²
27 a²+1 = 0 (mod 8)
27 a² = -1 (mod 8)
27 a² = 7 (mod 8)
3 a² = 7 (mod 8)
a² = 5 (mod 8)

but a² must be 0 or 1 (mod 8)

If you exchange the first and third members of the sequence you get:

27 a²-1 = 8 b²
27 a²-1 = 0 (mod 8)
27 a² = 1 (mod 8)
3 a² = 1 (mod 8)
a² = 3 (mod 8)

so it is also invalid.

Since 3² = 1 (mod 8), the number 27 cannot be replaced by 27*3^2k.

[mfgoode]

Originally Posted by wblipp

alpertron:

I don't see why your approach would prevent
the four consecutive numbers from being
27*a[sup]2[/sup]
8*b[sup]2[/sup]
c[sup]2[/sup]
6*d[sup]2[/sup]

:smile:
None of the above answers are convincing enough. Personally I would prefer a more rigorous proof, a real hard boiled one, with no ambiguity or speculation.

HINT: let the product be P, then in all cases it is one short of a perfect square Thus P + 1 = a perfect square and I leave it up to all to prove this

I will wait overnight and then will give a rigorous proof tomorrow.

Mally

[fetofs]

Originally Posted by mfgoode

:smile:
None of the above answers are convincing enough. Personally I would prefer a more rigorous proof, a real hard boiled one, with no ambiguity or speculation.

HINT: let the product be P, then in all cases it is one short of a perfect square Thus P + 1 = a perfect square and I leave it up to all to prove this

I will wait overnight and then will give a rigorous proof tomorrow.

Mally

May I refer you to Post #6?

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

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