An Interesting Diophantine Equation

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[@kaslircribs5804]

A Beautiful match between Math and Classical Music. Thank you!

[@christophebosquilloncrypto4555]

That was fun :) Follow up exercise: determine that, for any value of n>=6, the equation is divisible by 5 or 10 for any n being even or odd respectively. Get well soon!

[@zanti4132]

If n is a multiple of 5, the equation definitely isn't divisible by 5. Your statement is true if n is *not* divisible by 5, as in this case n⁴ = 1 mod 5.

[@MrGeorge1896]

Not for n = 15, 45, 75, etc... p is not divisible by 2, 3 or 5 for n = 30k - 15

[@amadeus-1011]

bachs french suite 2 . Love it ❤

[@geoffseyon3264]

What was the significance of the list 5…9 at the end?

[@chaosredefined3834]

Incomplete. n^4 - 20n^2 + 4 = (n^2 - 4n - 2)(n^2 + 4n - 2) You showed no integer solutions to get n^2 - 4n - 2 = 1 or n^2 +4n - 2 = 1. However, what if n^2 - 4n - 2 = -1 or n^2 + 4n - 2 = -1. Well... n^2 - 4n + 4 = 5 or n^2 + 4n + 4 = 5. So, (n - 2)^2 or (n + 2)^2 = 5. In both cases, 5 is not a perfect square, so no solution. Still no solution, but you should cover that case, just in case.

[@mukhibhai]

Can this be approached by noting that the LHS is always even, while the RHS odd (except the trivial 2 off course)?

[@yuryp6975]

LHS is odd if n is odd

[@zanti4132]

If n is even, then the LHS is a multiple of 4. That eliminates all the even integers.

[@scottleung9587]

I also got no integer solutions for n, given p is prime.

[@bobbyheffley4955]

The equation has been shown to have no integer solutions.

[@lucasjackson172]

why would it be prime???...

[@adamrussell658]

The question was _could_ it be prime.

[@rakenzarnsworld2]

No integer solutions...

[@glebyakovlev1321]

n is odd, given. n^2 = 2k + 1, much simpler

[@zanti4132]

Letting n = 2k + 1 establishes that p must be 1 more than a multiple of 4. Then what? p = 5, 13, 17, 29, etc. haven't been ruled out.

[@glebyakovlev1321]

@@zanti4132 not n, n^2

[@zanti4132]

@@glebyakovlev1321 Either way... unless I'm missing something, all you're showing is that p has to be one more than a multiple of 4. Now, it's not hard to prove n has to be a multiple of 5. If n is *not* a multiple of 5, then n⁴ = 1 mod 5, hence n⁴ - 20n² + 4 is divisible by 5. So n has to be an odd multiple of 5, which means n⁴ - 20n² + 4 has to equal 9 mod 10. It's down to where p has to be a prime ending in 9 that is one more than a multiple of 4: 29, 89, 109, 149, 229, etc. The approach shown in the video is cleaner.