Determine if $n^2-21n+111$ is or is not a perfect square (Second Solution ?)

Determine if $n^2-21n+111$ is or is not a perfect square.

Answer:

I want you to consider the following method of attacking the problem, and digest it, then think about it long enough so you have found the answer and explain back to me why it cannot deem to be a solution.

I first treat $n^2-21n+111$ as a square, says $m^2$ and I then rewrite $n^2-21n+111$ in the following fashion:

$n^2-21n+111=m^2$

$4(n^2-21n+111)=4m^2$

$4n^2-84n+444=4m^2$

$(2n-21)^2-21^2+444=4m^2$

$(2n-21)^2-3=4m^2$

$(2n-21)^2-4m^2=3$

$(2n-21+4m)(2n-21-4m)=3(1)\stackrel{\text{or}}=1(3)$

Solving $2n-21+4m=3$ and $2n-21-4m=1$ gives $m=1$, $n=10$ or $n=11$.

Solving $2n-21+4m=1$ and $2n-21-4m=3$ also leads to $m=1$, $n=10$ or $n=11$.

Therefore, $n^2-21n+111$ is a perfect square only at $n=10$ and $n=11$.

As usual, I will let you think about it for some time before I post the explanation some time today or tomorrow.