Any product of four consecutive integers is one less than a perfect square.(Hint: Take four consecutive integers as ) , m , m , mm +1 +2 +3
m(m+1)(m+2)(m+3) = (m^2+3m)^2 + 2(m^2+3m) = (m^2+3m + 1)^2 - 1
To solve this problem, let's assume that the four consecutive integers are m, m+1, m+2, and m+3.
We are given that their product is one less than a perfect square, which means we need to find a value for m that satisfies this condition.
To express the product of these four integers, we can multiply them: m * (m+1) * (m+2) * (m+3).
Now, let's simplify this expression:
m * (m+1) * (m+2) * (m+3) = (m^2 + 3m + 2m + 6) * (m^2 + 3m)
Expanding this expression further, we get:
(m^2 + 5m + 6) * (m^2 + 3m)
Multiplying these two binomials, we get:
m^4 + 8m^3 + 15m^2 + 18m
Now, we need to find the perfect square that is one less than this expression.
A perfect square can be represented as (m^2 + n)^2, where n is a constant.
So, we have:
(m^2 + n)^2 = m^4 + 8m^3 + 15m^2 + 18m - 1
Expanding this expression, we get:
m^4 + 2n * m^2 + n^2 = m^4 + 8m^3 + 15m^2 + 18m - 1
Now, we can compare the coefficients of m^3, m^2, and m from both sides of the equation:
Coefficient of m^3: 0 = 8
Coefficient of m^2: 2n = 15
Coefficient of m: 0 = 18
From the above equations, we can determine that the coefficient of m^3 is not equal to 8, which means there is no value of m that satisfies the condition.
Therefore, there are no four consecutive integers where their product is one less than a perfect square.