Determine if $n^2-21n+111$ is or is not a perfect square (Why the second solution fails?)

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

Answer:

Previously I wanted you to find the answer and explain back to me why the following solution cannot be deemed to be a sound 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 $n=11$."

The main reason why this won't work is because the original problem does not mention anything about $n$!

If it is to ask us to determine if $n^2-21n+111$ is or is not a perfect square for positive integer $n$, then the above solution above is sound and correct. But nope, we're not told that $n$ is a positive integer, so $n$ can be any real rational value!

In that case, we cannot of course to factor $3$ as in $(2n-21+4m)(2n-21-4m)=3$ to take ONLY the forms

$(2n-21+4m)(2n-21-4m)=3\cdot 1$ or $(2n-21+4m)(2n-21-4m)=1\cdot 3$

There is no way to prove that those two are the only solutions, we have to broaden our way to factor $3$, which in turn suggest a whole lot more cases to consider, one of those is

$(2n-21+4m)(2n-21-4m)=9\cdot \dfrac{1}{3}$

and we have to ascertain if the resulting $m$ and $n$ satisfy the condition that $n^2-21n+111$ is a perfect square.

Okay, once you have come to realize this, you also have to say goodbye to this method of attacking the problem, simply because there is no way to continue with it to look for all possible ways of factorizing $3$ to check if $n^2-21n+111$ is a perfect square.