Whole-Number Results from a Square Root

[TorontoTeacher]

Hello;

I've been working with rational equations, and this square root is involved:

\(\displaystyle {\sqrt{d^2 + 4 d + 36}}\)

The values of d that give whole-number values for this square root are -9, -4, 0 and 5. Why?

The only thing I've discovered so far is that the decimal portion of \(\displaystyle {\sqrt{d^2 + 4 d + 36}}\) seems to be symmetrically distributed around \(\displaystyle d=-2\).

 

[soroban]

Hello, TorontoTeacher!

\(\displaystyle \text{Given: }\:\sqrt{d^2 + 4 d + 36}}\)

\(\displaystyle \text{The values of }d\text{ that give integer values for this square root}\)
. . \(\displaystyle \text{are: }\,\text{-}9, \text{-}4, 0 \text{ and }5.\;\text{ Why?}\)

We want: .\(\displaystyle d^2 + 4d + 36 \:=\:k^2\)

Then:.\(\displaystyle d^2 + 4d + 4 + 32 \:=\:k^2 \quad\Rightarrow\quad (d+2)^2 + 32 \:=\:k^2\)

. . Hence, we have: .\(\displaystyle k^2 - (d+2)^2 \:=\:32\)

This is a Pellian equation and can be solved as such,
. . but I used a very primitive approach.


We have two squares that differ by 32.

Fact: Two squares will always differ by a sum of consecutive odd integers.

Can we partition 32 into a set of consecutive odd integers?

Yes, it can be done in exactly two ways: .\(\displaystyle \begin{Bmatrix}15,17 \\ 5,7,9,11\end{Bmatrix}\)

Hence, the squares are: .\(\displaystyle \begin{Bmatrix}9^2 - 7^2 & [1] \\ 6^2 - 2^2 & [2]\end{Bmatrix}\)

In case [1], we have: .\(\displaystyle k\,=\,\pm9,\;d+2\,=\,\pm7 \quad\Rightarrow\quad \boxed{d \:=\:5,\,\text{-}9}\)

In case [2], we have: .\(\displaystyle k\,=\,\pm6,\;d+2\,=\,\pm2 \quad\Rightarrow\quad \boxed{d \:=\:0,\,\text{-}4}\)

 

[TorontoTeacher]

Amazing reply.Before posting, I got to the equation you showed, but didn't know about "Pellian equations". I had a feeling it had to do with basic number theory.The beauty of your solution is not lost on me, and I've bookmarked the wikipedia page for Pell's Equation and I'll be reading about it later.Thanks very much. I stayed up later than I should have last night trying to figure it out