But it diverge to infinity!

Number Theory Level 3

$1! + 2! +3! + \ldots + x! = y^2$

Find the number of positive integer solutions $(x,y)$ such the the equation above holds true.

Inspired by RMO 2005 Problem.

4 1 2 3

2 solutions

Prakhar Bindal
Nov 8, 2015

Concept Used - Factorial Of Numbers Greater Than Or Equal To 5 Always End With A Zero .

If We Notice Carefully If x = 4 then 1!+2!+3!+4! = 33 . Now After This All Factorials Will End With A Zero And Hence There Sum Will End With A Three. No Perfect Square Ends With A Three . So No Solution Will Exist For x greater than or equal to 4 .

Hence Now We Can Check For x = 1,2,3 .

We Get Two Pairs (x,y) = (1,1) and (3,3)

bhai maine toh graph bana liya tha ! :)

A Former Brilliant Member - 4 years, 7 months ago

I don't think that $(x,y) = (3,3)$ satisfies the given equation, since $1! + 2! + 3! \ne 3!.$

Thus I believe that the answer should be $1$ and not $2.$

Brian Charlesworth - 5 years, 7 months ago

I suppose that he meant this - http://illumine.xyz/discussions/topic/if-1-2-3-x-y2-find-the-number-of-possible-ordered-pair-of-x-y/

Brahmnoor Singh - 5 years, 7 months ago

Sir I Added y^2 only to the original problem . but then i got a mail from brilliant saying "we have edited the problem kindly check " . and they have edited it wrongly!

Prakhar Bindal - 5 years, 7 months ago

Ah, o.k.. I wonder why they did that? With $y^{2}$ in place of $y!$ the answer of $2$ is indeed correct. :)

Brian Charlesworth - 5 years, 7 months ago

Anubhav Tyagi
Jun 18, 2016

GREAT LOGIC BEHIND A GREAT PROBLEM

But it diverge to infinity!

Inspired by RMO 2005 Problem.

2 solutions

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