Alright, so I know that this is a couple months late, but here it is.

In this note, I will discuss how to find the maximum possible integral value of both x and y such that \(\sqrt{a^2x^2+bx+c}\) is an integer (note that \(b\neq 2a\sqrt{c}\) or else it can be factored to \((ax+c)^2\) which is integral for all integral values of ax).

Begin with the generic quadratic \(y=\sqrt{a^2x^2+bx+c}\).

Since \(a^2x^2\) is guaranteed a perfect square, \(bx+c\) must be a difference of squares. A short proof for this is \(y^2=a^2x^2+bx+c\Rightarrow y^2-a^2x^2=bx+c\)

This means that the left side can be rearranged to

\[\begin{eqnarray}
ax+n &=& \sqrt{a^2x^2+bx+c}\\
a^2x^2+2axn+n^2 &=& a^2x^2+bx+c\\
bx-2axn&=&n^2-c\\
x(b-2an)&=&n^2-c\\
x&=&\boxed{\dfrac{n^2-c}{b-2an}}\\
\end{eqnarray}\]

NOTE: \(n\) is a variable that is an integer.

This is maximized when the denominator is minimized or as the numerator tends infinity. Thus to minimize the denominator, \(n\approx \dfrac{b}{2a}\) (look familiar? It's the same as the negation of the vertex of a parabola. Why this is, I have no idea). The \(\approx\) sign (it means approximately) is important for two reasons. 1) because \(n\) has to be integral for y to be integral and \(\dfrac{b}{2a}\) isn't always integral. 2) \(n\) can't exactly equal \(\dfrac{b}{2a}\) because if it does, then we are dividing by 0 and the world will implode.

NOTE: this is a plug and chug because we need to find an integral value of n that will make the numerator divisible by the denominator.

Now that we have found the maximum value of x, we can plug this in to find y.

\(y=\sqrt{a^2x^2+bx+c}\)

\(y=\sqrt{a^2\left(\dfrac{n^2-c}{b-2an}\right)+b\left(\dfrac{n^2-c}{b-2an}\right)+c}\)

\(rearranging...\)

\(\boxed{y=\left|\dfrac{bn-a(n^2+c)}{b-2an}\right|}\)

(the long bars denote absolute value.. sorry it looks so bad)

This also further proves that y is maximized when \(n\) is as close as possible to \(\dfrac{b}{2a}\).

## Practice section

Maximum possible integer square root polynomial 1 (easier)

Maximum possible integer square root polynomial 2 (harder)

## Comments

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TopNewestI have a few questions:

Why did you let \(y=ax+n\)?

Why is \(f(n)=\dfrac{n^2-c}{b-2an}\) maximized at \(n\approx \dfrac{b}{2a}\)?

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The first is easy to answer. If we look at the second half of the \(a^2x^2+bx+c\), then since the first half is already a perfect square, n basically represents the difference between squares. Thus it can be positive or negative. If b and c are positive, it will be positive, if both are negative, n is negative. If one is negative and one is positive, then it gets complicated.

For The second half, it is not maximized there. But that is the largest possible integral value that is achievable by that function. I don't have a "full" proof as I'm very bad at explaining this kind of thing. If you'd like me to I could try to type up a full explicit proof for this.

Which also reminds me. I have a proof of why n must be integral for x to be integral. I could add that if you'd like me to.

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How did you wrote \[y =ax+n\]?

And there is \[x^{2}\] missing in the same expression.

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I don't know what you mean by missing a \(x^2\)

But \(y=ax+n\) because \(a^2x^2\) is guaranteed a perfect square, so \(bx+c\) must be a difference between two squares. Assuming that \(bx+c\neq 0\), n is just an arbitrary value that is the difference between \(a^2x^2\) and \(y^2\). You could guess many numbers for the Max value of x to find a perfect square value for \(a^2x^2+bx+c\), but you would never know when to stop. This method allows you to check for values of n and know when you find the max value for x (since the denominator should be minimized).

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there it is

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