Number Theory question

[Steven G]

I tried everything I know about number theory ( a course which I never took before) but can not solve the problem below. Can someone please solve it for me or give a leading hint? Thanks!

Prove that there exist two distinct natural numbers m and n with m, n≤2049 such that 19^m−19ⁿ is divisible by 2019.The truth of this claim does not depend on the exact values 19, 2019, and 2049. So, your solution should give an argument that is easy to adapt to other values for these constants rather than, say, using a computer program to find specific values for m and n.

 

[Deleted member 4993]

I tried everything I know about number theory ( a course which I never took before) but can not solve the problem below. Can someone please solve it for me or give a leading hint? Thanks!

Prove that there exist two distinct natural numbers m and n with m, n≤2049 such that 19^m−19ⁿ is divisible by 2019.The truth of this claim does not depend on the exact values 19, 2019, and 2049. So, your solution should give an argument that is easy to adapt to other values for these constants rather than, say, using a computer program to find specific values for m and n.

2019 is a prime number.

So for the conjecture to be true:

19^n is divisible by 2019 ......... and/or

[19^(m-n) -1] is divisible by 2019

It is always true that:

The unit at number for 19^(A) is either 9 [for odd A] or 1 [for even A]

19 and 2019 are relatively prime.

continue...

 

[Steven G]

2019 is a prime number.

So for the conjecture to be true:

19^n is divisible by 2019 ......... and/or

[19^(m-n) -1] is divisible by 2019

It is always true that:

The unit at number for 19^(A) is either 9 [for odd A] or 1 [for even A]

19 and 2019 are relatively prime.

continue...

Honestly, that is exactly how far I got. May I get another hint?

 

[Deleted member 4993]

Honestly, that is exactly how far I got. May I get another hint?

Why did not you tell us about your journey in the OP?

2019 cannot divide into 19^m !

So the culprit must be the other factor ... . Continue...

 

[Steven G]

Why did not you tell us about your journey in the OP?

2019 cannot divide into 19^m !

So the culprit must be the other factor ... . Continue...

Yes! So [19^(m-n) -1] is divisible by 2019.
I then said 19^(m-n) = 1 mod 2019. This is honestly how far I got. WA gave me the solution but when I explain this to my student I can't use software.