12^n + 25^n+1 is divisible by 13
Let us assume that P(n) is the statement "12n + 25n+1 is divisible by 13"
subs. n=1 , P(1) :
12+ 252 = 637 = 13x49
... P(1) is true
Now, let us assume that P(k) is true for all n=k,
=> 12k + 25k+1 = 13A , where A is any suitable constant.
=> 12k = 13A - 25k+1
Let us prove that P(k+1) is also true by using P(k)
P(k+1):
12k+1 + 25(k+1)+1 = 12.12k + 25.25k+1
12. (13A - 25k+1) + 25.25k+1
13(12A) - 12(25k+1) + 25(25k+1)
= 13(12A + 25k+1)
=13B , where B = 12A + 25k+1
Hence P(n) is true for all 12n + 25n+1
subs. n=1 , P(1) :
12+ 252 = 637 = 13x49
... P(1) is true
Now, let us assume that P(k) is true for all n=k,
=> 12k + 25k+1 = 13A , where A is any suitable constant.
=> 12k = 13A - 25k+1
Let us prove that P(k+1) is also true by using P(k)
P(k+1):
12k+1 + 25(k+1)+1 = 12.12k + 25.25k+1
12. (13A - 25k+1) + 25.25k+1
13(12A) - 12(25k+1) + 25(25k+1)
= 13(12A + 25k+1)
=13B , where B = 12A + 25k+1
Hence P(n) is true for all 12n + 25n+1