AIME Problem: sum of all n>0 where n^2-19n is square

[Trenters4325]

Find the sum of all positive integers n for which n^2 -19*n +99 is a perfect square.

 

[soroban]

Re: AIME Problem

Hello, Trenters4325!

Find the sum of all positive integers $\displaystyle n$ for which $\displaystyle n^2\,-\,19n\,+\,99$ is a perfect square.

Let: $\displaystyle \,n^2\,-\,19n\,+\,99\;=\;k^2$

Then we have the quadratic equation: $\displaystyle \,n^2\,-\,19n\,+\,(99\,-\,k^2)\;=\:0$

Quadratic Formula: $\displaystyle \:n\;=\;\frac{-(-19)\,\pm\,\sqrt{(-19)\,-\,4(1)(99\,-\,k^2)}}{2(1)} \;= \;\frac{19\,\pm\,\sqrt{4k^2\,-\,35}}{2}$

Since $\displaystyle n$ is an integer, $\displaystyle 4k^2\,-\,35$ must be a square, say, $\displaystyle a^2.$

Then we have: $\displaystyle \,4k^2\,-\,35\:=\:a^2\;\;\Rightarrow\;\;(2k)^2\,-\,a^2\:=\:35$

We have two squares that differ by 35.

The only such squares are: $\displaystyle \,(6^2,\,1^2)$ and $\displaystyle (18^2,\,17^2)$
$\displaystyle \;\;$That is: $\displaystyle \,k\,=\,3$ or $\displaystyle k\,=\,9$

If $\displaystyle k\,=\,3,\;n\:=\:\frac{19\,\pm\,1}{2}\:=\:10,\,9$

If $\displaystyle k\,=\,9:\;n\:=\:\frac{19\,\pm\,17}{2}\;=\;18,\,1$


The sum of the $\displaystyle n$'s is: $\displaystyle \,1\,+\,9\,+\,10\,+\,18\;=\;38$

 

[Trenters4325]

Re: AIME Problem

We have two squares that differ by 35.

The only such squares are: $\displaystyle \,(6^2,\,1^2)$ and $\displaystyle (18^2,\,17^2)$

$\displaystyle \;\;$That is: $\displaystyle \,k\,=\,3$ or $\displaystyle k\,=\,9$

Did you just come up with those pairs mentally?

 

[soroban]

Re: AIME Problem

Hello, Trenters4325!

Did you just come up with those pairs mentally?

No, I had some help . . .

Consecutive squares differ by consecutive odd numbers.

. . $\displaystyle 0^2\underbrace{\qquad\quad}1^2\underbrace{\qquad\quad}2^2\underbrace{\qquad\quad}3^2\underbrace{\qquad\quad}4^2\underbrace{\qquad\quad}5^2\underbrace{\qquad\quad}6^2\underbrace{\qquad\quad}\qquad\cdots$
. . . . .$\displaystyle 1\qquad\qquad3\qquad\qquad5\qquad\qquad7\qquad\qquad9\qquad\qquad11\qquad\cdots$

If the difference is an odd number, $\displaystyle 2n\,+\,1$,
the smaller square is $\displaystyle n^2$; the larger is $\displaystyle (n\,+\,1)^2$.

We have the difference: $\displaystyle \,2n\,+\,1\;=\;35\;\;\Rightarrow\;\;n\,=\,17$

And that's how I found: $\displaystyle 18^2\,-\,17^2\:=\:35$


Sometimes there are more solutions.

Since $\displaystyle \,35\;=\;5\,\times\,7\;=\;7\,+\,7\,+\,7\,+\,7\,+\,7$
$\displaystyle \;\;$which can be modified to: $\displaystyle \,3\,+\,5\,+\,7\,+\,9\,+\,11$

and since these are the differences of consecutive squares, we have:
. . . . .$\displaystyle 3\qquad\qquad\:5\qquad\qquad\:7\qquad\qquad\:9\qquad\qquad\:11$
. . $\displaystyle \;\,\overbrace{\qquad\qquad}\;\,\overbrace{\qquad\qquad}\;\,\overbrace{\qquad\qquad}\;\;\overbrace{\qquad\qquad}\;\;\overbrace{\qquad\qquad}$
. . $\displaystyle 1^2\qquad\qquad2^2\qquad\qquad3^2\qquad\qquad4^2\qquad\qquad5^2\qquad\qquad6^2$

And that's how I found: $\displaystyle 6^2\,-\,1^2\:=\:35$