1. **State the problem:** Prove that for any non-negative integer $n$, the expression $3^{2n} + 7$ is divisible by 8. 2. **Formula and approach:** We want to show that $3^{2n} + 7 \equiv 0 \pmod{8}$, meaning $3^{2n} + 7$ leaves a remainder of 0 when divided by 8. 3. **Evaluate powers modulo 8:** Let's find the pattern of $3^{2n} \pmod{8}$. - Calculate $3^2 \pmod{8}$: $$3^2 = 9 \equiv 1 \pmod{8}$$ - Since $3^2 \equiv 1 \pmod{8}$, then: $$3^{2n} = (3^2)^n \equiv 1^n = 1 \pmod{8}$$ 4. **Substitute back:** $$3^{2n} + 7 \equiv 1 + 7 = 8 \equiv 0 \pmod{8}$$ 5. **Conclusion:** Since $3^{2n} + 7$ is congruent to 0 modulo 8, it is divisible by 8 for all non-negative integers $n$. **Final answer:** $3^{2n} + 7$ is divisible by 8 for all $n \geq 0$.