Mathcounts National 2014 Sprint #23

Question

I have been attempting this problem for some time now and can't seem to figure it out. If someone could help I would be grateful.

My attempt:

$4b^2 + 4b + 1 = n^2$
$3b^2 + 5b + 1 = (n-2)^2$
$b^2 - b = n^2 - (n^2+4-4n) = 4n-4$
$b(b-1) = 4(n-1)$
This led me to believe that b = n = 4. However, for this to be true, $4(16) + 4(4) + 1$ would have to equal $4^2$ .
When this failed, I backtracked.
$b^2-b=4n-4$
$(b-2)^2 = 4n$
This let me find out that n is a square number, and it is positive. I tried substitution various squares but could not find anything that fit the original two equations:
$4b^2 + 4b + 1 = n^2$
$3b^2 + 5b + 1 = (n-2)^2$

Answer 1

From $4b^2 + 4b + 1 = n^2$, you should be able to conclude that $n = 2b + 1$.

Putting it into the equation $3b^2 + 5b + 1 = (n - 2)^2$, you should get $b^2 - 9b = 0$ and conclude that $b = 9$. And $n = 2b + 1 = 19$.