Stirling Approximation : Solve a^11+b^11=11!

Using Stirling approximation, let us try to solve the equation $a^{11}+b^{11}=11!$ and find the behavior of the values of a and b under various conditions, and what restrictions apply on a and b when using Stirling’s approximation. The concepts involved in this problem solution are concepts of Diophantine Equations, Elementary Number theory, Inequalities,Stirling Approximation and Exponential Diophantine equations. Here we not only discuss the solution conditions for the above problem but also consider whether the above equation solution yields representation of a Diophantine Equation?

Stirling Approximation problem results in a Diophantine equation?

The problem to be proved here is as in the image below.

Let us try to understand what is Diophantine equation and Diophantine exponential Equation. Here $a^{11}+b^{11}=11!$ can be a Diophantine equation, if we get the integer solution for the above problem for a and b. Let us try to apply the Stirling approximation technique to find out whether we can get the real integer solution for a and b.

Concept of Stirling Approximation

The sterling approximation is given by the sterling formula as below to approximate the value of a large factorial like 11!.

$n!\approx \sqrt{2\mathit{\pi }n}(\frac{n}{e}{)}^n$

Where n is any integer and n!= Factorial of n.

Here we have to find the value of 11!. Hence n=11.

From Stirling Approximation,

$11!\approx \sqrt{2\mathit{\pi }\times 11}(\frac{11}{e}{)}^{11}$ , hence,

$(11{\mathrm{!)}}^{1/11}\approx (\sqrt{22\mathit{\pi }}{)}^{1/11}\times \frac{11}{e}\approx 4.91$

Application of Stirling Approximation

But from the given problem, $a^{11}+b^{11}=11!$ ,

$a^{11}<11!$ and $b^{11}<11!$ ,

From above derivations,

$a<(11{\mathrm{!)}}^{1/11}$ and $b<(11{\mathrm{!)}}^{1/11}$ , hence,

$a,b<(11{\mathrm{!)}}^{1/11}\approx 4.91$

Condition for validity of Diophantine Equation

If Diophantine Equation conditions are met, then a and b should be integers.

Hence from $a,b<(11{\mathrm{!)}}^{1/11}\approx 4.91$ ,

The above is condition for the given problem to represent a Diophantine Equation if a,b are also integers.

Checking for integer solutions for a and b

From above the largest values of a and b can be 4.

Hence $a^{11}+b^{11}=4^{11}+4^{11}=8388608$

But without calculating 11! we can say that the last two digits of 11! = 00 as 10 and 5,2 are factors of 11!.

Hence $a^{11}+b^{11}\ne 11!$

Hence the given problem can not have integers solutions. for a and b.

As above the given problem does not have integers solutions for a and b is proved. The value of factorial on right hand side is more and more large, then the Stirling approximation will become the best method to solve that type of problems.

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Conclusion

As above, the concepts of Stirling approximation , Diophantine equation which are part of Elementary Number theory, Inequalities pertaining to use and solving the equations involving the exponential Diophantine equations. A Diophantine equation can be defined as the combination of variables as integers. If the variables cannot take integer values then the test of Diophantine equation fails. Here we could establish and prove that the solution of above problem does not yield and integer values for a and b. To prove this, we have taken the help of Diophantine equation concepts and Stirling approximation.

best of luck to every one.