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Old 2020-10-21, 17:52   #1
petrw1
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Default Can someone help me understand this...

Or simplify it:

How does such a seemingly odd string of irrational numbers equate to exactly 142?
Better yet how would someone have come up with this --- mathematically; not via trial and error?

Interestingly, the first half is almost exactly twice the second half.

(√√7! + √√7!) * (√(√7! + (√7!)/7!)) = 142

Thanks
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Old 2020-10-21, 18:25   #2
Viliam Furik
 
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The whole process is here.

If we say the 7 factorial is x, then we can generalize it for x, and by shifting things around, we find that expression of the form (√√x + √√x) * √(√x + (√x)/x) is a whole number whenever x+1 is a square.
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Old 2020-10-21, 18:48   #3
petrw1
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Quote:
Originally Posted by Viliam Furik View Post
The whole process is here.

If we say the 7 factorial is x, then we can generalize it for x, and by shifting things around, we find that expression of the form (√√x + √√x) * √(√x + (√x)/x) is a whole number whenever x+1 is a square.
Thanks a lot....no big deal, but you dropped he 2x on line 2 of your napkin.

So If I knew what I was doing I could reverse this process and get other whole numbers starting with √(x!+1) ??
x can be 4, 5, or 7.

Last fiddled with by petrw1 on 2020-10-21 at 18:54
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Old 2020-10-21, 19:26   #4
Viliam Furik
 
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Quote:
Originally Posted by petrw1 View Post
Thanks a lot....no big deal, but you dropped he 2x on line 2 of your napkin.
Yes, I know, thus the implication arrow afterwards, instead of an equality sign. Removing the 2 doesn't change the rest. If I were to get some fraction in the end, I could simply put it back. But it doesn't change the rationality, which is the whole question.

And, it isn't a napkin. (But I think you know that) It is my notebook I use for maths in the school. (BTW, I am graduating in May 2021, at least that's what I thought a year ago. Who knows what else might Covid take - we are learning online since the last Monday, and it's possible it will be until Christmas)

Quote:
Originally Posted by petrw1 View Post
So If I knew what I was doing I could reverse this process and get other whole numbers starting with √(x!+1) ??
x can be 4, 5, or 7.
Yes! Absolutely. You can shove in any number for x, even primorials, perfect powers, and also Riesel primes with even powers of base and square ks (based on few look-ups, they might not exist), but sadly enough, no Mersenne primes except M2, as 2p - 1 + 1 = 2p, which is not a square if the p is odd.

----
EDIT:
Silly me. Of course there can't be a Riesel prime with k being square and n being even, because of the almighty algebraic factors of a2 - 1 = (a-1)(a+1)

Last fiddled with by Viliam Furik on 2020-10-21 at 19:36
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Old 2020-10-22, 07:43   #5
LaurV
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Quote:
Originally Posted by petrw1 View Post
(√√7! + √√7!) * (√(√7! + (√7!)/7!)) = 142
Let \(\alpha=\sqrt{7!}\). You have \(2\sqrt\alpha\cdot\sqrt{\alpha+\frac{\alpha}{\alpha^2}}\). Which, when multiply the radicals and simplify the fraction under it, becomes \(2\sqrt{\alpha^2+1}\). Now substitute back the \(\alpha\), you have \(2\sqrt{7!+1}\), or \(2\sqrt{5041}\), which is 2*71.
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Old 2020-10-22, 11:37   #6
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Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives

2x^{\frac{1}{4}}\cdot\(x^{\frac{1}{2}}\;+\;x^{\frac{-1}{2}}\)^{\frac{1}{2}}

The "obvious" multiplication then gives 2\sqrt{x+1}.

I note that things can go wrong for complex values of "x" that aren't positive real numbers.
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Old 2020-10-22, 16:45   #7
petrw1
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Thanks all
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Old 2020-10-27, 17:45   #8
petrw1
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I can follow the simplification that gets one to 2 x SQRT(7!+1) = 142

Can someone explain how you would start at 2 x SQRT(7!+1) = 142
and given that the final equation must contain only 7's, get to the original equation I supplied in post 1.
For example the 2 at the front could become something like (7+7)/7 OR <any function with a 7> + <same> iff that function can be divided out in another part of the entire formula to equal 2.
....confused??? Me too.

(My son has a game/puzzle app when he must find a formula using the least number of each digit from 1 to 9 to get a number)

He is limited to +. -, x, /, SQRT and concatenation. ie 77 or 777
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Old 2020-10-28, 04:54   #9
LaurV
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Do you mean like engineeers112?.
(scroll through the ppt presentation there)
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Old 2020-10-28, 05:42   #10
petrw1
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Quote:
Originally Posted by LaurV View Post
Do you mean like engineeers112?.
(scroll through the ppt presentation there)
Exactly ... that is sooooo much clearer!!!!
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