what should be added to product of four consecutive numbers to make it perfect square number.

Question

It is related to number system and basic concepts of number theory

Answer 1

??????? with 4 konsequit numbers ?????????

zampel...1*2*3*4=24

next square=25, so add 1

Answer 2

1. Let the product of 4 consecutive integers be n(n+1)(n+2)(n+3), n:every integer.

2. Let the integer that is added to the product be 'A'.

3. Let the number squared that makes the sum of the procuct and 'A' be (n^2 +an +b)^2, a,b: integer.

So that, the question can be restated as follows:

n(n+1)(n+2)(n+3) + A = (n^2 + an + b)^2 ··· Eq.1

Here, n(n+1)(n+2)(n+3) + A = n^4 + 6n^3 + 11n^2 + 6n + A ··· Eq.2

(n^2 + an + b)^2=n^4 +2an^3 + (2b + a^2)n^2 + 2abn + b^2 ··· Eq.3

Compare coefficients of the like terms in Eq.2 and Eq.3.

2a=6, 2b + a^2=11, 2ab=6, so a=3, b=1 and b^2=1, that is: A=b^2=1 

Eq.1 is restated: n(n+1)(n+2)(n+3) + 1 = (n^2 +3n +1)^2

CK: If n=11, 11x12x13x14+1=24024+1=24025, (11^2 + 3x11 +1)^2=(155)^2=24025, so LHS=RHS

If n=-3, (-3)(-2)(-1)(0)+1=1, ((-3)^2 + 3(-3) +1)^2=(1)^2, so LHS=RHS   CKD.

The answer: 1 is added to the product of four consecutive numbers to make the sum a squared number.