Show that 11^(n + 2) + 12^(2n + 1) is divisible by 133 for every positive integer n, and find the…

Step 1. Rewrite. $11^{n + 2} + 12^{2n + 1} = 121 \times 11^n + 12 \times (133 + 11)^n$.

Step 2. Expand. $(133 + 11)^n = 11^n + 133 \times (\text{an integer})$.

Step 3. Combine. The total is $121 \times 11^n + 12 \times 11^n + 12 \times 133 \times (\text{an integer}) = 133 \times 11^n + 133 \times (12 \times \text{an integer})$.

Step 4. Conclude. Every term has a factor $133$.

For $n = 2$. $11^4 + 12^5 = 14641 + 248832 = 263473 = 133 \times 1981$.