Prove that 25+8*(n!)) is not square

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Goahead52
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Prove that 25+8*(n!)) is not square

Postby Goahead52 » Wed Jul 08, 2015 6:02 pm UTC

Hi,

I have hard time proving that :

k=25+8*(n!) is not a perfect square.
Any idea?

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Re: Prove that 25+8*(n!)) is not square

Postby gmalivuk » Wed Jul 08, 2015 7:18 pm UTC

Why do you expect that to be true in the first place? And why is it important?
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Goahead52
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Re: Prove that 25+8*(n!)) is not square

Postby Goahead52 » Wed Jul 08, 2015 7:26 pm UTC

I have found that for some numbers like 3 for example.
If you start by adding each time 1 to the last number and summing you will never reach some factorial.

(3+4+5+6+..............+k) will never be equal to some factorial.

I hope to be wrong.

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Re: Prove that 25+8*(n!)) is not square

Postby >-) » Wed Jul 08, 2015 11:04 pm UTC

Somewhat related to this https://en.wikipedia.org/wiki/Brocard%27s_problem

I tested up to n = 20000 without finding any squares

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Re: Prove that 25+8*(n!)) is not square

Postby quintopia » Thu Jul 09, 2015 8:20 am UTC

That is not just somewhat related. It is a huge step in the problem. That link points to a proof that the abc conjecture implies the OP's problem has finitely many solutions. From potentially infinite to provisionally finite is like... Halfway to zero, right?

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Re: Prove that 25+8*(n!)) is not square

Postby ZeroPoint » Tue Jul 21, 2015 1:29 am UTC

Not got anything proved, but think I have a form that may be easier to deal with. We want to have a^2 =25 +8 n!. Then (a-5)(a+5) =8n!. Now make the transform to b=a-5. Now b(b+10)=8n!. This implies 2|b or b+10, and so 2|b. Now set k=b/2. Then k(k+5)=2n!. In particular, we now know we must group all powers of 2 together. Just thought id chip in - but this may not be helpful

Goahead52
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Re: Prove that 25+8*(n!)) is not square

Postby Goahead52 » Sat Jul 25, 2015 1:22 pm UTC

Little up.
Maybe someone have found a new approach maybe not.

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Re: Prove that 25+8*(n!)) is not square

Postby lorb » Wed Jul 29, 2015 6:34 pm UTC

if n>=25 then 25+8*(n!) is always divisible by 25. So if it is a perfect square it has to be the square of a number divisible by 5.
(For n<25 we already know there are no squares by previous poster)
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Re: Prove that 25+8*(n!)) is not square

Postby Demki » Wed Jul 29, 2015 7:41 pm UTC

lorb wrote:if n>=25 then 25+8*(n!) is always divisible by 25. So if it is a perfect square it has to be the square of a number divisible by 5.
(For n<25 we already know there are no squares by previous poster)

That's actually true for n>=10. Since 10! Is divisible by 25.

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Gwydion
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Re: Prove that 25+8*(n!)) is not square

Postby Gwydion » Fri Jul 31, 2015 5:25 am UTC

Demki wrote:
lorb wrote:if n>=25 then 25+8*(n!) is always divisible by 25. So if it is a perfect square it has to be the square of a number divisible by 5.
(For n<25 we already know there are no squares by previous poster)

That's actually true for n>=10. Since 10! Is divisible by 25.

More specifically an odd multiple of 5 which is not also divisible by 3, since 25+8n! is odd and equivalent to 1 mod 3 for n>2.

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Re: Prove that 25+8*(n!)) is not square

Postby Cradarc » Sun Aug 09, 2015 6:34 am UTC

25 + 8(n!) = y2
Let y = 10nk+5, where gcf(k,10) = 2

Do some algebra to obtain:
8(n!) = (10n+1k)(10n-1k + 1)
n! = 25b(b+1)/2, where b = 10n-1k. Alternatively, n! = 25*T(b), where T(i) = ith triangle number

In order for the equation to be satisfied, any prime less than n>10 must divide exclusively b or (b+1).
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