# 12^n + 25^n+1 is divisible by 13

^{n }+ 25

^{n+1 }is divisible by 13"

subs. n=1 , P(1) :

12+ 25

^{2 }

^{ }= 637 = 13x49

_{.}

^{.}

_{. }P(1) is true

Now, let us assume that P(k) is true for all n=k,

=> 12

^{k }+ 25

^{k+1 }= 13A , where A is any suitable constant.

=> 12

^{k }= 13A - 25

^{k+1}

Let us prove that P(k+1) is also true by using P(k)

P(k+1):

12

^{k+1 }+ 25

^{(k+1)+1 }= 12.12

^{k }+ 25.25

^{k+1}

12. (13A - 25

^{k+1}) + 25.25

^{k+1}

13(12A) - 12(25

^{k+1}) + 25(25

^{k+1})

= 13(12A + 25

^{k+1})

=13B , where B = 12A + 25

^{k+1 }

Hence P(n) is true for all 12

^{n}+ 25

^{n+1}

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