Question

if the product of four consecutive natural numbers increased by a natural number 'P' it is a perfect square then find the value of 'P'

Answer 1

Let us say the numbers are n, n+1, n+2 & n+3. We need the value of "p" which will make {n(n+1)(n+2)(n+3)+p} a perfect square.

Let us consider the product n(n+1)(n+2)(n+3)
= n(n+3)(n+1)(n+2) ..... just rearranged
= (n^2 + 3n) * (n^2 + 3n + 2)
= a * (a +2) ...... where a = n^2 + 3n
= a^2 + 2a

Now we know that since n is a natural number, n^2+3n will be a natural number and hence a will be a natural number as well.

Therefore the product now is in the form a^2 + 2a clearly (1) short of a perfect square.
If p=1, then the sum of the product and p becomes, a^2 + 2a + 1 = (a+1)^2.

Hence, the required value of p = 1.

Hope it helps

Answer 2

Answer:

four consecutive number is x(x+1)(x+2)(x+3)

And than answer is p=1