prove that for all integral n, An=11^(n+2)+12^(2n+1) is divisible by 133

This can be done with some algebra gymnastics.

Prove 11^{n+2}+12^{2n+1}=133p, for some integer p.

Verify n=1 is true: 11^{3}+12^{3}=3059, \;\ 3059/133=23... TRUE.

Assume P(k)=11^{k+2}+12^{2k+1}=133p is true.

We have to show that P(k+1) is true.

Rewrite:

P(k+1)=11^{k+3}+12^{2k+3}=11\cdot{11^{k+2}}+144\cd ot{12^{2k+1}}

Add and subtract 133\cdot{11^{k+2}}:

11\cdot{11^{k+2}}+133\cdot{11^{k+2}}+144\cdot{12^{ 2k+1}}-133\cdot{11^{k+2}}

=144\cdot{11^{k+2}}+144\cdot{12^{2k+1}}-133\cdot{11^{k+2}}

=144(\underbrace{11^{k+2}+12^{2k+1}}_{\text{this is 133p}})-133\cdot{11^{k+2}}

=144(133p)-133\cdot{11^{k+2}}

=\underbrace{133(144p-11^{k+2})}_{\text{multiple of 133}}

Therefore P(k+1) is true and the induction is complete. QED.

prove that for all integral n, An=11^(n+2)+12^(2n+1) is divisible by 133
Thanks!!