Find the smallest of them all

Algebra Level 3

For some positive integers $n$ , $n^2 - 2016$ is a perfect square. Find the smallest possible value for $n$ .

The answer is 45.

2 solutions

Chew-Seong Cheong
Feb 21, 2017

Let the perfect square be $(n-m)^2$ , then

$\begin{aligned} (n-m)^2 & = n^2 - 2016 \\ n^2 - 2mn + m^2 & = n^2 - 2016 \\ -2mn + m^2 & = -2016 \\ \implies n & = \frac {2016+m^2}{2m} \\ & = \frac {1008}m + \frac m2 & \small \color{#3D99F6} \text{Using AM-GM inequality} \\ n & \ge 2 \sqrt{504} \approx 44.900 \end{aligned}$

Therefore, the smallest $n$ is $\boxed{45}$ and $45^2 - 2016 = 9 = 3^2$ .

Note: We could have directly concluded that $n^2 - 2016 \geq 0 \Rightarrow n \geq \sqrt{2016} = 2 \sqrt{ 504}$ .

Unfortunately, this approach isn't great because it is very unique to the value of 2016. For example, if we had $( n- m) ^2 = n^2 - 2017$ , then we would still arrive at $n \geq 45$ , and then start testing the cases $n = 45, 46, \ldots$ . Unfortunately, it would take us a long time to get to the answer of $n = 1009$ .

Calvin Lin Staff - 4 years, 3 months ago

The 2017 version does not require much testing. By using the standard method (same as mine and possibly others) and notation of the solution of Tom Engelsman and the fact, that 2017 is a prime number, hence the only possible factorisation is:

$n - k = 1$

$n + k = 2017$

$2n = 2018$

$n = \boxed {1009}$

Zee Ell - 4 years, 3 months ago

This is why I said "Unfortunately, this approach isn't great because it is very unique to the value of 2016. "

To be clear, I'm referring to Chew-Seong's solution, where he finds the minimium value of $n$ and then tests this value. I am not referring to "all possible approaches to this problem".

Calvin Lin Staff - 4 years, 3 months ago

beautiful. great minds and experience can give such a unique solution like this. mine was a standard method but was also easy

Sathvik Acharya - 4 years, 3 months ago

Tom Engelsman
Feb 21, 2017

Let $n^2 - k^2 = 2016 \Rightarrow (n-k)(n+k) = 2016.$ Now $2016 = 2^{5}3^{2}7^{1}$ , which contains eighteen positive integer divisor pairs:

$(n-k, n+k) = (1, 2016); (2, 1008); (3, 672); (4, 504); (6, 336); (8, 252); (9, 224); (12, 168); (14, 144); (16, 126); (18, 112); (21, 96); (24, 84); (28, 72); (32, 63); (36, 56); (42, 48).$

Of these pairs, $(n-k, n+k) = (42, 48)$ yields the smallest $n:$

$n - k = 42$

$n + k = 48$

or $2n = 90 \Rightarrow \boxed{n = 45}.$

Find the smallest of them all

The answer is 45.

2 solutions

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