Find all Integer Solutions - A Modular Arithmetic Trick

We find all integer solutions to the equation 6^m - 5^n = 11. This involves a trick using modular arithmetic, which we explore in more depth at the end.

00:00 Intro
00:50 Solution
04:22 Choice of modulo k
07:50 Euler's theorem
09:54 Summary

[GaetanoDiCaprio]

Very nice solution, but the extraordinary value of this video is the last part: an in-depth analysis of the heuristics underlying the resolution process. Wonderful!

[pableraspfgpfg]

Hi @DrBarker,
According to this logic, why not using mod 12? It shares also the same prime factors than 6, (so it will get 6^m mod 12 to be 0 even before) and it also will get 5^n mod 12 to loop between 1 and 5, making it quite easy to spot the solution.
Is there any reason I am not seeing?

[tidadmathsci]

Thanks for the mod trick. One can also find the solution with a basic method.
Notice that 11 = 1*11 = 6^m - 5^n = (6^(m/2))^2 - (5^(n/2))^2 = (6^(m/2) - 5^(n/2))(6^(m/2) + 5^(n/2)) the only possibility is
6^(m/2) - 5^(n/2) = 1 and 6^(m/2) + 5^(n/2) = 11 which can be solved easily to get m=2 and n=2

[primsiren]

I had success attacking modulo 7. I don't really know anything about number theory so at first I just randomly tried mod 5 but that wasn't useful at all (I don't think taking mod of any of the bases is useful, so mod 5, 6 and 11 wouldn't be useful) then I tried mod 7 which works well.

6^m ~ (-1)^m mod7
5^n ~ (-2)^n mod 7
both of these can be seen from binomial expansion or recursion like you did in the video to see the pattern.

11 ~ 4 mod 7
or
11 ~ -3mod 7

there are 4 cases:

case 1: -1 - (-2)^n =4
no sols (parity)
case 2: 1- (-2)^n =4
no sols (parity)
case 3: -1 - (-2)^n =-3
=> (-2)^n = -2
no sols
case 4: 1 - (-2)^n =-3
=> (-2)^n =4
=> n=2
a quick check reveals m=2 and of course this is the only solution.

This makes me wonder whether in general it is a good idea to take modulo of the (highest base + 1) as it ensures the +- 1 mod (highest base + 1) on one term.

For example for the general a^m - b^n = L where L, a, b are positive integers. Take mod(a+1)
=> a^m ~ (-1)^m mod(a+1)
suppose b<a
suppose also that b+c = a+1, where c is an integer, after binomial expansion I get b^n ~ (-c)^n mod(a+1)
=> (-1)^m - (-c)^n = L mod(a+1)
which may or may not have solutions (?) but it should be finite anyway so you can check all cases.

Please tell me what you think and please make some more videos on number theory (especially abusing modular arithmetic).

[MyOneFiftiethOfADollar]

The modular truths are necessary but not sufficient.

conjecture: k should have all the prime divisors of one of the bases.

Will make up a problem and see if the conjecture is useful like it was here where 6 and 24 have the same prime factors.

[AT-zrtv]

Neat trick! Would you have the time to consider generalizing the approach to solving / attempting to solve a^m - b^n = c, where (a, b, c) is given?

[mcwulf]

So I started mod 6 and figured n must be even, -(-1)^n == -1 (mod 6)
I then used mod 25 to knock out 5^n and 6^m == 1, 6, 11, 16, 21, 1, 6... etc (mod 25). As we want 6^m = 11 then m=2 also. I didn't get round to proving it was the only solution!

Would never have chosen mod 24!

[holyshit]

m=n=2
This is my guess before watching
There are probably no more solutions