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Solve the equation in real numbers.

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Solve the equation in real numbers.

Postby Math Tutor » Sun Mar 03, 2013 11:23 pm

Solve the equation in real numbers.
1! + 2! +3! + ... + x!=y^2


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Re: Solve the equation in real numbers.

Postby Guest » Mon Mar 04, 2013 9:37 am

I'm assuming you mean x is a positive integer and y is an integer.
(The sum 1!+2!+3!+\ldots+x! is ambiguous when x is not a positive integer,
and if y is allowed to be a real then y = \pm \sqrt{1!+2!+3!+\ldots +x!} are the solutions)

Assuming x and y are positive integers the solution is given below.
Spoiler: show
If x\geq 4, then 1!+2!+3!+\ldots +x! \equiv 1!+2!+3!+4! \equiv 3 \bmod 5
y can always be written in the form y = 5n+r for some integer n and r = 0,1,2,3,4 (simply divide y by 5, and take the quotient as n and the remainder as r).
This means y^2 = 25n^2 + 10nr + r^2 \equiv r^2 \bmod 5. No value of r gives y^2\equiv 3 \bmod 5, so y^2\ne 1!+2!+3!+\ldots +x! when x \geq 4.

When x\leq 3, there are two solutions x=1,3 and y=1,3 respectively.


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