Don't Try Small Cases

Number Theory Level 4

For what positive integer $n$ is

$n^2 + 13n + 3$

a perfect square ?

Bonus: Can you prove that the answer is unique?

The answer is 33.

1 solution

Brian Charlesworth
Jan 24, 2017

Let $m$ be a positive integer such that $n^{2} + 13n + 3 = m^{2}$ . Then by completing the square we have

$\left(n + \dfrac{13}{2}\right)^{2} - \left(\dfrac{13}{2}\right)^{2} + 3 = m^{2} \Longrightarrow (2n + 13)^{2} - 13^{2} + 3*4 = 4m^{2} \Longrightarrow$

$(2n + 13)^{2} - (2m)^{2} = 157 \Longrightarrow (2n + 13 + 2m)(2n + 13 - 2m) = 157$ .

Now since $157$ is prime and $n,m$ positive, we must then have that

$2n + 13 + 2m = 157$ and $2n + 13 - 2m = 1$ , and so

$(2n + 13 + 2m) + (2n + 13 - 2m) = 157 + 1 \Longrightarrow 4n + 26 = 158 \Longrightarrow \boxed{n = 33}$ .

Note that $33$ is not unique over all integers. $n = -46$ also makes the given expression a perfect square.

I was playing around with these and trying to find a "large unique" solution, which happens when that constant (157) is large and prime.

Chung Kevin - 4 years, 4 months ago

Yes, it is kind of fun to play around with finding $a,b$ so that $n^{2} + an + b = m^{2}$ , (A), has a unique solution. I find that if we can find $a,b$ so that $a^{2} - 4b$ is prime then (A) will have the unique positive solution $n = \dfrac{(a - 1)^{2}}{4} - b$ . Having $a$ odd will ensure that $n$ will be an integer, but after that we have to play around with $b$ to make $a^{2} - 4b$ prime. Some examples where $b$ is minimized for a given odd $a$ :

$n^{2} + 3n + 1 \Longrightarrow a^{2} - 4b = 5, n = 0$ ,
$n^{2} + 5n + 2 \Longrightarrow a^{2} - 4b = 17, n = 2$ ,
$n^{2} + 7n + 2 \Longrightarrow a^{2} - 4b = 41, n = 7$ ,
$n^{2} + 9n + 2 \Longrightarrow a^{2} - 4b = 73, n = 14$ ,
$n^{2} + 11n + 2 \Longrightarrow a^{2} - 4b = 113, n = 28$ ,
$n^{2} + 13n + 3 \Longrightarrow a^{2} - 4b = 157, n = 33$ ,
$n^{2} + 15n + 7 \Longrightarrow a^{2} - 4b = 197, n = 42$ ,
$n^{2} +17n + 2 \Longrightarrow a^{2} - 4b = 281, n = 62$ ,
$n^{2} + 19n + 2 \Longrightarrow a^{2} - 4b = 353, n = 79$ ,
$n^{2} + 21n + 2 \Longrightarrow a^{2} - 4b = 433, n = 98$ .

O.k., I'm getting carried away here, but I was just looking for patterns. The only ones I'm seeing are that $b = 2$ is a predominant minimum and $n$ being a multiple of $7$ is disproportionally common.

Brian Charlesworth - 4 years, 4 months ago

I came to the same conclusion in your first paragraph :)
I used a table of primes to determine when $a^2 - 4b$ was prime.

I didn't make the observation that $b = 2$ was very common.
Since $a^2 - 4 = (a-2)(a+2)$ , $b=1$ will be prime only for $a = 3$ . I can believe that $b = 2$ is common especially for small $a$ . How does this compare for larger values of $a$ , especially when the prime gaps start to get larger?

Chung Kevin - 4 years, 4 months ago

@Chung Kevin – I agree, one would expect a minimum $b = 2$ to become less common as $a$ increases. The series $(a,b,n)$ continues with

$(23,2,119), (25,2,142), (27,5,164), (29,3,193), (31,2,223),$

$(33,5,251), (35,2,287), (37,2,322), (39,7,354),$

$(41,3,397), (43,12,429), (45,2,482), (47,12,517), ...$

So while $2$ does keep showing up, it appears do so increasingly less frequently as $a$ increases. The divisibility of $n$ by $7$ seems to be becoming less frequent as well. $n$ has been prime $7$ times out of $23$ so far, which is right in line with the $14$ primes occurring in the first $46$ positive integers, so there doesn't seem to be any patterns there.

Brian Charlesworth - 4 years, 4 months ago

Sir is there a proof that only 33 is the positive integral solution to the above problem

Achal Jain - 4 years, 4 months ago

Because $157$ is prime and $m,n$ are positive, the only way to break down the factors is to set $2n + 13 + 2m = 157$ and $2n + 13 - 2m = 1$ , which leads to $n = 33$ as the only solution.

Brian Charlesworth - 4 years, 4 months ago

More accurately, because $2n + 13 + 2m$ is positive, and $157 = (2n+13+2m)(2n+13-2m)$ , hence we need only consider the positive factors of 157. Since $2n + 13 + 2m > 2n + 13 - 2m$ , and 157 is prime, hence we must have $2n = 13 + 2m = 147, 2n + 13 - 2m = 1$ .

Note that $2n+13 - 2m$ could possibly be negative. Furthemore, if we allowed $n$ to be negative, then $2n + 13 + 2m$ could be negative. In this case, we have to look at $157 = -1 \times -157$ , which gives us $2n + 13 +2m = -1, 2n + 13 - 2m = -157$ , or that $n = -46$ as Brian pointed out at the end of his solution.

Calvin Lin Staff - 4 years, 4 months ago

Blegh I went wrong because I threw away the square at my by accident...

Peter van der Linden - 4 years, 4 months ago

Don't Try Small Cases

The answer is 33.

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