(7^m- 2^m) is odd as well as (7^m - 2^m) is odd. So, any one of
m,n can not be 2 or 7. since 2^m mod 7 cannot be 0.
Case(i): Let m,n take values other than 5.
7^m = 7 mod m (by Fermat's Little Theorem)
2^m = 2 mod m.
7^m - 2^m = 5 mod m. So, can not be divisble by m and therefore
must be divisible by n. Similarly, 7^n - 2^n must be
divisible by m.
WOLG take m>n
Then, 7^m-2^m mod n =0
=> 7^m = 2^m mod n => (7*(1/2))^m mod n =1
7*(1/2) should be relatively prime to n as remainder is 1.
Let the least integer for which (7*(1/2))^x mod n =1 is c, then m mod c=0.
=> m = k*c,where c is a divisor of n-1(Fermat's little theorem),
but m is a prime, so either k=1 or c=1
If k=1, m=c this violates m>n.
If c=1 => then 7*(1/2) should be 1 which is impossible
becasue only value for which x*(1/2)modn=1
is x=2(Existence of unique inverse for a field).
So, there are no solutions for this case.
case(ii): Atleast one of m,n =5
m=n=5 is obviously the solution because 7^5-2^5 = 5 mod 5 = 0 mod5
Consider 7^5-2^5.
All the prime positive divisors of this number will be
solutions of the given equation.
7^5-2^5=25*11*61. So, solutions are as follows:
(5,5),(5,11) and (5,61).
Here T(x): Euler's Totient function.
Edited on August 31, 2007, 5:55 am
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Posted by Praneeth
on 2007-08-31 01:58:15 |
Solution
