Given the following quadratic equations:
Is it possible to determine how many values of n will generate a perfect square? Or better yet, is it possible to determine which values of n will generate a perfect square?
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HINT: Let's rearrange the first equation:
$4n^2-4n+132n+1-132=(2n-1)^2+132(n-1)=m^2$ now we can easily see that $n-1$ has to be $0$ so $n=1$
For the second equation we may write:
$4n^2+4n+12n+1-12=(2n+1)^2+12(n-1)=m^2 \Rightarrow n=1$
For the third equation:
$4n^2+20n+4n+25-28=(2n+5)^2+4(n-7)=m^2 \Rightarrow n=7$
You can complete the square. For the third example, $4n^2+24n-3=(2n+6)^2-39.$ As $(n+1)^2-n^2=2n+1$, this cannot be a perfect square for any $2n+6$ greater than $20$, leaving only $7$ possibilities.
We can do better. You want $(2n+6)^2-39=p^2.$ This means $39=(2n+6-p)(2n+6+p)$. There will be one solution for each way to factor $39$ into two numbers. Because of the $2n,$ you need the sum of the numbers to be even, which will be true for odd numbers like $39$ and will be the case for your other three examples.