Hi,

I have hard time proving that :

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

Any idea?

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

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

Why do you expect that to be true in the first place? And why is it important?

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

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.

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.

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

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

I tested up to n = 20000 without finding any squares

I tested up to n = 20000 without finding any squares

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

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?

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

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

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

Little up.

Maybe someone have found a new approach maybe not.

Maybe someone have found a new approach maybe not.

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

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)

(For n<25 we already know there are no squares by previous poster)

Please be gracious in judging my english. (I am not a native speaker/writer.)

http://decodedarfur.org/

http://decodedarfur.org/

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

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.

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

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.

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

25 + 8(n!) = y

Let y = 10

Do some algebra to obtain:

8(n!) = (10

n! = 25b(b+1)/2, where b = 10

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

^{2}Let y = 10

^{n}k+5, where gcf(k,10) = 2Do some algebra to obtain:

8(n!) = (10

^{n+1}k)(10^{n-1}k + 1)n! = 25b(b+1)/2, where b = 10

^{n-1}k. Alternatively, n! = 25*T(b), where T(i) = ith triangle numberIn order for the equation to be satisfied, any prime less than n>10 must divide exclusively b or (b+1).

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