Solving Brocard's Problem: What Values for n?

[Ashwin_Kumar]

Hello, recently i have been dwelling on the topic of brocard's problem(alongside some summation problems). I'm sure most of you are familiar with the problem: if n!+1 is a square number, what values other than 4,5 and 7 can n be? So far i have only extended the problem in an attempt to reduce it to a previously solved one, to no avail. I'm currently
\frac{\pi\alpha}{\Gamma(1-\alpha)cos(\frac{1}{2}-\alpha)}+1=x^2

in which alpha is the number we are trying to find. If anyone knows how to continue from here, please tell me. Thanks.

 

[Ashwin_Kumar]

Sorry, there was a slight problem with the equation. it should be

\frac{\alpha\pi}{\Gamma(1-\alpha)sin(\alpha\pi)}+1=x^{2}

 

[Ashwin_Kumar]

If anyone knows anything that could be used, please post it on this thread.

 

[dimension10]

Sorry, there was a slight problem with the equation. it should be

\frac{\alpha\pi}{\Gamma(1-\alpha)sin(\alpha\pi)}+1=x^{2}

sin(a*pi)=0

Thus the answer is infinity.

However, Gamma(1-a) is meaningless. Thus the question itself makes no sense.

 

[Mute]

sin(a*pi)=0

Thus the answer is infinity.

However, Gamma(1-a) is meaningless. Thus the question itself makes no sense.

\Gamma(1-a) is not meaningless. It has a finite value whenever 1-a is not a negative integer or zero, and when 1-a is a negative integer or zero, it diverges. It's perfectly well defined, and the fact that it diverges when 1 - a is a negative integer means that \Gamma(1-a)\sin (\pi a) is an indeterminate form, and in this case will give a finite answer. Using the reflection formula, \Gamma(1 + \alpha) = \pi/(\Gamma(-\alpha)\sin (\pi \alpha)) = -\alpha \pi/(\Gamma(1-\alpha)\sin(\pi\alpha)).

To the OP: I think you've made the task harder for yourself by using the reflection formula. You would probably be better of staying with the expression

\Gamma(1 + \alpha) + 1 = x^2
as \alpha only appears once on the left hand side of the expression. This would, in principle, make it easier to solve for \alpha as \alpha(x) = \Gamma^{-1}(x^2-1)-1, where \Gamma^{-1}(z) is the "inverse Gamma function". However, I am not aware of any references to such a function, so it may not exist or be well defined. Looking at a graph of the real part of the gamma function, it may be well defined for real z > 1, where the Gamma function is monotonic.

One could then plot alpha vs. x, and see for what values of x alpha is an integer.

 

[dimension10]

\Gamma(1-a) is not meaningless. It has a finite value whenever 1-a is not a negative integer or zero, and when 1-a is a negative integer or zero, it diverges. It's perfectly well defined, and the fact that it diverges when 1 - a is a negative integer

Yes, but 1-alpha is negative all the time when alpha is positive.

 

[dimension10]

To the OP: I think you've made the task harder for yourself by using the reflection formula.

Or does it? You cannot really get much from knowing that Γ(1+α)+1=x2 unless you use the reflection formula.

 

[Mute]

Yes, but 1-alpha is negative all the time when alpha is positive.

Two things, just to be clear: \Gamma(z) is well defined for all complex z, except where z is zero or a negative integer, where it diverges. However, even if a is always a postive integer, such that \Gamma(1-a) diverges, one has to remember that it is multiplied by \sin(\pi a) in this expression, which is zero whenever the Gamma function is infinite. As you can see by plotting \Gamma(1-x)\sin(\pi x), this function is continuous and has a well defined limit when x is an integer, so \Gamma(1-x)\sin(\pi x) can be interpreted as its limit values when x is an integer (just as sinc(x) is 1 at x = 0 and not undefined). This is the way it must be interpreted, as the factor of -\alpha\pi/(\Gamma(1-\alpha)\sin(\pi\alpha)) came from using the reflection formula on \Gamma(1+\alpha), which is perfectly well defined to begin with.

Or does it? You cannot really get much from knowing that Γ(1+α)+1=x2 unless you use the reflection formula.

How does the reflection formula help? The OP essentially wants to know what values of alpha will give him a number that is a perfect square. If he could solve for alpha as a function of x, he could compute \alpha(x) for integers x to see which ones result in alpha being an integer. The reflection formula just introduces more factors of alpha, making it harder to solve for alpha. I really don't see how it helps solve the OP's problem.

 

[dimension10]

Two things, just to be clear: \Gamma(z) is well defined for all complex z, except where z is zero or a negative integer, where it diverges. However, even if a is always a postive integer, such that \Gamma(1-a) diverges, one has to remember that it is multiplied by \sin(\pi a) in this expression, which is zero whenever the Gamma function is infinite. As you can see by plotting \Gamma(1-x)\sin(\pi x), this function is continuous and has a well defined limit when x is an integer, so \Gamma(1-x)\sin(\pi x) can be interpreted as its limit values when x is an integer (just as sinc(x) is 1 at x = 0 and not undefined). This is the way it must be interpreted, as the factor of -\alpha\pi/(\Gamma(1-\alpha)\sin(\pi\alpha)) came from using the reflection formula on \Gamma(1+\alpha), which is perfectly well defined to begin with.

I guess you are right, but the answer would be infinity because the sine of any multiple of pi radians is 0.

 

[dimension10]

How does the reflection formula help? The OP essentially wants to know what values of alpha will give him a number that is a perfect square. If he could solve for alpha as a function of x, he could compute \alpha(x) for integers x to see which ones result in alpha being an integer. The reflection formula just introduces more factors of alpha, making it harder to solve for alpha. I really don't see how it helps solve the OP's problem.

Well, by using the reflection formula, we see that there is no other number because the sine of alpha*pi is 0 and thus the answer would be infinite?

 

[Mute]

I guess you are right, but the answer would be infinity because the sine of any multiple of pi radians is 0.

No, the answer is not infinity. As a tends to an integer, the expression \Gamma(1-a)\sin(\pi a) tends to the indeterminate form \infty \dot 0, which can turn out to be 0, infinity, or a finite value. In this case, the limit is a final value. If you haven't studied this yet, please see this wikipedia article.

Well, by using the reflection formula, we see that there is no other number because the sine of alpha*pi is 0 and thus the answer would be infinite?

But we already know there are solutions: n! + 1 gives perfect squares for at least 4, 5 and 7. The OP is trying to find out if there are more values of n which give perfect squares. So, you know your argument has a flaw already, because looking at the reflection formula there's no indication that 4, 5 or 7 are special values.

 

[dimension10]

No, the answer is not infinity. As a tends to an integer, the expression \Gamma(1-a)\sin(\pi a) tends to the indeterminate form \infty \dot 0, which can turn out to be 0, infinity, or a finite value. In this case, the limit is a final value. If you haven't studied this yet, please see this wikipedia article.

I do know that infinity multiplied by 0 is the interdeterminate form since any number divided by infinity is 0. However, when I posted my earlier post, I did not see why Γ(1−a) is infinity. I thought that by Euler's reflection formula, a negative number can have a finite factorial.
Γ(1−a) Γ(a)=pi/sin(pi*a)
However, now I do understand that the answer is not infinity because the sine of any multiple of pi is 0 and thus the answer is 0*infinity

 

[dimension10]

Now, I doubt it is even possible...

 

[seywondenigma]

i think i found some solutions with microsoft excel program though... i don't know how right they are so i am still examining them..