x-2-y-1-x-z-0-y-1-y-z-find-x-N-such-that-equation-is-true-

Question Number 826 by 123456 last updated on 18/Mar/15

x^{2} = \underset{y = 1}{\sum^{x}} \underset{z = 0}{\prod^{y - 1}} (y - z)

f i n d x \in N^{*} s u c h t h a t e q u a t i o n i s

t r u e

Commented by 123456 last updated on 18/Mar/15

\underset{z = 0}{\prod^{y - 1}} (y - z)

z = 0 \Rightarrow y - z = y

z = y - 1 \Rightarrow y - z = 1

\underset{z = 0}{\prod^{y - 1}} (y - z) = y!

\underset{y = 1}{\sum^{x}} \underset{z = 0}{\prod^{y - 1}} (y - z) = \underset{y = 1}{\sum^{x}} y! = 1! + 2! + \cdot \cdot \cdot + x!

x^{2} = 1! + \cdot \cdot \cdot + x!

Answered by prakash jain last updated on 18/Mar/15

x = 3 and x = 1 are solutions .

x! > x^{2} for x ⩾ 4

x = 4, x! = 24, x^{2} = 16, x! > x^{2}

If x! > x^{2} then for x > 4

\Rightarrow (x + 1)! = x! (x + 1) > x^{2} (x + 1) = x^{2} + x^{3}

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

> x^{2} + 1 + x^{2} + x + 1 > x^{2} + 1 + x + x + 1 ∵ x^{2} > x \land x > 1

> {(x + 1)}^{2} + 1 > {(x + 1)}^{2}

So there is no solution for x ⩾ 4 .