MHB Can $(n^2+11n-4) \cdot n! + 33 \cdot 13^n + 4$ Be a Perfect Square?

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The discussion revolves around the problem of determining for which positive integers \( n \) the expression \( (n^2+11n-4) \cdot n! + 33 \cdot 13^n + 4 \) results in a perfect square. The poster encourages participation by extending the time for responses, noting that previous problems went unanswered. An official solution is provided for those interested in the mathematical approach. The community is invited to engage with the problem, highlighting the challenge and complexity involved. Ultimately, the goal is to find specific values of \( n \) that satisfy the condition of the expression being a perfect square.
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Here is this week's POTW:

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Find all positive integers $n$ such that $(n^2+11n-4)\cdot n!+33\cdot 13^n+4$ is a perfect square.

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Hello MHB! I have decided to give the community another week's time to try to attempt last week's POTW! (Nod)
 
No one answered last two week's POTW. (Sadface) However, for those who are interested, you can read the official solution below:
Let us denote $a_n=(n^2+11n-4)\cdot n!+33\cdot 13^n+4$. If $n\ge 4$, then 8 divides $n!$. Hence, $a_n\equiv 33\cdot 13^n+4\equiv 5^n+4 \pmod 8$. As $5^2\equiv 1 \pmod 8$, then $5^2\equiv 1 \pmod 8$ for all even $n$. Therefore, $a_n\equiv 5 \pmod 8$, if $n\ge 4$ and $n$ is even. But perfect squares leave remainders 0, 1 or 4 when dividing by 8.

Secondly, when $n\ge 7$, then 7 divides $n!$. So $a_n\equiv 33 \cdot 13^n+4\equiv 5\cdot (-1)^n+4 5 \pmod 7$. Therefore, for odd $n\ge 7$, $a_n\equiv =-5+4=-1 \pmod 7$. But perfect squares leave remainders 0, 1, 4 or 2 when dividing by 7.

We are left with possible candidates $n=1,\,2,\,3$ and $n=5$. For $n=5$, $a_n$ is not a perfect square because $a_5=33\cdot 13^5+4\equiv 3^6-1\equiv 3 \pmod 5$, but perfect squares leave remainder 0, 1 or 4 when divided by 5. Also, $a_n$ is not a perfect ssquare for $n=3$ because $a_3=(9+33-4)\cdot 6+33\cdot 13^3+4\equiv 3+3^4-1\equiv 3\pmod 5$. Finally, we can check that $a_1=(1+11-4)\cdot 1+33\cdot 13+4=441=21^2$ and $a_2=(4+22-4)\cdot 2+33\cdot 169+4=5625=75^2$.
 
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