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Problem Of The Week #467 May 10th 2021

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,909
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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,909
Hello MHB! I have decided to give the community another week's time to try to attempt last week's POTW! (Nod)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,909
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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