Croatian Mathematical Olympiad

### Claim. For every positive integer $n\ge 3$ there exists a positive integer $k$ such that $51^{2^{n-2}}-1=2^n\cdot (2k+1)$.

Proof. We prove the claim by mathematical induction on $n$.

For $n=3$ we have $51^2-1=3^2-1=8$ (mod 16), so $51^2-1$ is divisible by $2^3$ but not by $2^4$, therefore the claim holds for $n=3$. Assume the claim holds for some $n$, i.e. $51^{2^{n-2}}-1=2^n\cdot (2k+1)$ for some positive integer $k$. Then $$\begin{array}{rl}51^{2^{n-1}}-1& =(51^{2^{n-2}}-1)(51^{2^{n-2}}+1)\\ & =2^n\cdot (2k+1)(2^n\cdot (2k+1)+2)\\ & =2^{n+1}\cdot (2k+1)(2^{n-1}(2k+1)+1),\end{array}$$ and since $(2k+1)(2^{n-1}(2k+1)+1)$ is odd, the claim holds for $n+1$. This completes the proof. $\square$

Regarding the initial statement of the problem, we will inductively define an appropriate $k_n$ for every $n$. Note that for $n=1,n=2$ and $n=3$ we can select $k=2$, i.e. $k_1=k_2=k_3=2$. Assume the claim holds for some $n$, i.e. there exists some positive integer $a_n$ such that $51^{k_n}-17=2^n\cdot a_n$. If $a_n$ is even, we can select $k_{n+1}=k_n$. If $a_n$ is odd, we can define $k_{n+1}=k_n+2^{n-2}$. Then we have $$51^{k_{n+1}}-17=51^{2^{n-2}}\cdot (51^{k_n}-17)+17\cdot (51^{2^{n-2}}-1).$$ By the inductive assumption and the previously proven claim, we can conclude that there exists a positive integer $k$ such that $$\begin{array}{rl}51^{k+n+1}-17& =51^{2^{n-2}}\cdot 2^n\cdot a_n+17\cdot 2^n\cdot (2k+1)\\ & =2^n\cdot (51^{2^{n-2}}{a}_n+17(2k+1)).\end{array}$$ Since $a_n$ is odd, it follows that $51^{2^{n-2}}{a}_n+17(2k+1)$ is even, therefore $51^{k+n+1}-17$ is indeed divisible by $2^{n+1}$.