Perfect Square Quadratic

write it as $n^2+24n+16=n^2+24n+144-128=(n+12)^2-2^7$ substitute $n+12=z$ , and let the whole expresion be $t^2$ $z^2 -2^7=t^2\longrightarrow (z-t)(z+t)=2^7$ now, lets say $z-t=y,z+t=x$ then $2z=x+y$ to attain the max of z, y must be as big as possible $(z-t)(z+t)=2\times 2^6\longrightarrow z-t=2,z+t=64$ $2z_{max}=66\longrightarrow z_{max}=33$ since $z=n+12$ $(n+12)_{max}=33\longrightarrow n_{max}=\boxed{21}$

If a nonnegative $n$ exists, then clearly it is larger than any negative value of $n$ . We assume $n$ is nonnegative.

Note that $(n+4)^2\leq n^2+24n+16<(n+12)^2$ .

Checking all possible perfect squares between the bound, we can easily get the largest possible integer $n$ .

For $n= 0$ we clearly have a perfect square ( $16$ ). So we can restrict our search among the positive integers.

Since $n$ is positive the expression $n^2 + 24n + 16$ is greater than $n^2$ so if it is perfect square, then it must be written as $(n+p)^2$ for some positive integer $p$ .

So

$n^2 + 24n + 16 = (n+p)^2$

leads to

$2n(p-12) = 16 - p^2$

Now you can easily note 3 interesting facts.

1) $p$ is even

(PROOF: $2$ divides $16 - p^2$ , and $16$ so $2$ divides also $p^2$ and (since $2$ is prime) $p$ must be even.)

2) $p$ is NOT a multiply of $3$

(PROOF: otherwise $3$ would divide $16 - p^2$ and $p^2$ and that should imply that $3$ divides $16$ )

3) $4 < p < 11$

(PROOF: because the two members of the equation must be either both negative or both positive (or both zero of course) ).

These 3 facts imply that $p \in \{ 8, 10 \}$ .

We can try these two values on the equation and from

$n = \frac{16 - p^2}{2(p-12)}$

we get two possible values for $n$ namely

$n = 6,\ \ \ \ n =21$

For $n = 6$ we get the initial expression to be the square of $n+p = 8 + 6 = 14$ .

For $n = 21$ we get the initial expression to be the square of $n + p = 21+10 = 31$ .

So the largest $n$ requested is $21$ .

Complete the square to get (n+12)^2-128=A^2

Where A is an integer.

Let S=n+12

S^2-128=A^2 (Equation 1)

Inspecting the sequence of square numbers 1,4,9.16,25......... We can see the differences are 3,5,7,9...... i.e. the increases increase by 2 each time.

We can write that for any square K^2 it has increased from the previous square number by 2K-1 (call this equation 2). e.g. 5^2=4^2+(2x5-1)

Therefore to maximise S (and hence n) we must find square numbers that have a difference of 128 (by looking at equation 1) and preferrably make them as close as possible to maximise.

Clearly all the diffferences between the squares are odd 3,5,7,9....... So S and A cannot be consecutive as their difference is 128 if they are squared. Therefore introduce a number Z between them so the number line looks like:

1,2,3.............A,Z,S.............

Let the difference between A^2 and Z^2 be denoted as D. Hence the difference between Z^2 and S^2 is D+2. As D and D+2 must add to 128, D=63 (2D+2=128).

A^2+63=Z^2

Using Equation 2 2Z-1=63 Z=32 Hence S=33 And therefore n=21 as S=n+12.

Let $n^{2} + 24n+16 = m^{2}$

Since $n \in I$ so $D=k^{2}$ where $k\in I$ , $D$ is the discriminant of the equation

$D=b^{2}-4ac =512+4m^{2}=k^{2}$

$\Rightarrow 512=(k-2m)(k+2m)$

Largest value of $m=31$

Substitute the value $m$ in the starting equation yo get

$n^{2}+24n-945=0$

$\huge (n-21)(n+45)=0$

$\Huge n_{max}=21$