Can someone help me understand this...

[petrw1]

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

[Viliam Furik]

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.

[petrw1]

Originally Posted by Viliam Furik

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.

[Viliam Furik]

Originally Posted by petrw1

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)

Originally Posted by petrw1

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 2^p - 1 + 1 = 2^p, 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)

[LaurV]

Originally Posted by petrw1

(√√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.

[Dr Sardonicus]

Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives



The "obvious" multiplication then gives .

I note that things can go wrong for complex values of "x" that aren't positive real numbers.

[petrw1]

Thanks all

[petrw1]

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

[LaurV]

Do you mean like engineeers112?.
(scroll through the ppt presentation there)

[petrw1]

Originally Posted by LaurV

Do you mean like engineeers112?.
(scroll through the ppt presentation there)

Exactly ... that is sooooo much clearer!!!!