Q.Prove by induction that $5^{2n} - 1$ is divisible by $24$ for every natural number $n$.

[!TLDR] Write $5^{2(k+1)}-1$ as $25(5^{2k}-1)+24$. [!ANSWER] Proved: $5^{2n}-1$ is divisibl …

Let $P(n)$ be the statement that $5^{2n}-1$ is divisible by 24. Base case: $n=1$: $5^2-1=25-1=24$, divisible by 24, so $P(1)$ holds. Inductive step: assume $P(k)$: $5^{2k}-1=24m$ for some integer $m$. Then $5^{2(k+1)}-1=5^{2k+2}-1=25\cdot5^{2k}-1=25(5^{2k}-1)+25-1=25(24m)+24=24(25m+1)$. Since $25m+1$ is an integer, $5^{2(k+1)}-1$ is a m …

Divisibility-by-induction: write $5^{2k+2}=25\cdot5^{2k}$ and …

Checking only that $5^{2n}-1$ is divisible by a smaller factor like 8 or 3 and stopping there, instead of verifying the full claimed …