Answer: For all non-negative integers \(n\), \(11^{2n}+25^n+22\) is divisible by \(24\).
Let \(f(n)=11^{2n}+25^n+22\). We will prove by induction that \(24\mid f(n)\) for all non-negative integers \(n\).
First, when \(n=0\),
\(f(0)=11^0+25^0+22=1+1+22=24\),
which is divisible by \(24\).
Now assume that for some non-negative integer \(k\), \(f(k)\) is divisible by \(24\). So we can write \(f(k)=24r\) for some integer \(r\).
Consider \(f(k+1)-f(k)\):
\(f(k+1)-f(k)=\bigl(11^{2k+2}+25^{k+1}+22\bigr)-\bigl(11^{2k}+25^k+22\bigr)\).
So
\(f(k+1)-f(k)=11^{2k}(11^2-1)+25^k(25-1)\)
\(=11^{2k}(121-1)+25^k(24)\)
\(=11^{2k}\cdot 120+25^k\cdot 24\)
\(=24\bigl(5\cdot 11^{2k}+25^k\bigr)\).
Thus \(f(k+1)-f(k)\) is divisible by \(24\). Since \(f(k)\) is also divisible by \(24\), their sum \(f(k+1)\) is divisible by \(24\).
Therefore, by mathematical induction, \(11^{2n}+25^n+22\) is divisible by \(24\) for all non-negative integers \(n\).
