Optionnal differences of squares.

Number Theory Level 3

How many ways can you make the equation true such that $x$ and $y$ are positive integers?

x^2-y^2=2016

6 12 24 32

4 solutions

Mateo Doucet De León
Mar 8, 2021

Method 1: all factors

Factors of 2016:

1	2	4	8	16	32
3	6	12	24	48	96
9	18	36	72	144	288
7	14	28	56	112	224
21	42	84	168	336	672
63	126	252	504	1008	2016

Subtract one for each number that is a multiple of 2, but not of 4 (2, 6, 10, 14...). For all other numbers, add one.

+	-	+	+	+	+
+	-	+	+	+	+
+	-	+	+	+	+
+	-	+	+	+	+
+	-	+	+	+	+
+	-	+	+	+	+

Total: 24

Options: total $/2=12$ solutions

Note: whenever the number of solutions is a non-integer, this means that the number you're factoring out is a square number and that the half-counted solution is when

y=0

Methed 2: prime numbers (recommended)

Define 2016 by its prime factors. Subtract one from the exponent of the base 2 and add one for the exponent of each of the other bases. Take each of the changed exponents and do the product between these numbers.

$2016=$	$\times 2^5$	$\times 3^2$	$\times 7^1$	Total
	-1	+1	+1	(product)
	4	3	2	24

Options: total $/2=12$ solutions

Proof

These products all equals 2016. Since 2016 is an even number, each have to be even as well.

$2\times 1008$	$4\times 504$	$8\times 252$	$16\times 126$
$6\times 336$	$12\times 168$	$24\times 84$	$48\times 42$
$14\times 144$	$28\times 72$	$56\times 36$	$112\times 18$

Each product can be turned into differences of squares knowing that $p\times q=(\frac{p+q}{2})^2-(\frac{p-q}{2})^2 (p\geq q)$ since $(a+b)(a-b)=a^2-b^2$

$505^2-503^2$	$254^2-250^2$	$130^2-122^2$	$71^2-55^2$
$171^2-165^2$	$90^2-78^2$	$54^2-30^2$	$45^2-3^2$
$79^2-65^2$	$50^2-22^2$	$46^2-10^2$	$65^2-47^2$

You have a typo in Method 1: "Subtract one for each number that is a multiple of 4, but not of 2. Else, add one" should be "for each number that is a multiple of 2, but not of 4".

It would also be good for Method 2 to clarify that "On each exponent, subtract one for 2" you mean to subtract one from the exponent of the base 2, not if the exponent is 2. As you show in your table 2^5 => exponent of base 2 is 5 => subtract one from exponent => 5-1 = 4. Before seeing the table, I interpreted it as "3^2 => exponent of base 3 is 2 => subtract one from exponent => 2-1 = 1.

Bill Chapp - 3 months ago

Thanks for helping, correction done.

Mateo Doucet De León - 3 months ago

Mateo, the biggest pain consists finding all the divisors, but here https://www2.math.upenn.edu/~deturck/m170/wk2/numdivisors.html there is a way to find them more easily

Lu Ca - 3 months ago

Iva Topuria
Mar 9, 2021

Let's N to be the number of ways we can get positive integer solutions for $x^{2} - y^{2} = 2016$ , n to be the number of divisors of 2016 and $n_{o}$ the number of odd divisors of 2016, so since $2016 = 2^{5}\times3^{2}\times7^{1} \Rightarrow n = (5 + 1)\times(2 + 1)\times(1 + 1) = 36$ .

Now we have to disregard the odd divisors $d_{oi}$ and their complements $c_{oi}$ such that $d_{oi}\times c_{oi} = 2016$ . The odd divisors of 2016 are 1, 3, 9, 7, 21 and 63, so $n_{o} = 6$ . Therefore, $N = \frac{n - 2n_{o}}{2} = \frac{36 - 2\times6}{2} = \boxed{12}$

Paul Romero - 3 months ago

S Broekhuis
Mar 9, 2021

$(Factor. pairs. found. using. https://www.calculatorsoup.com/calculators/math/factors.php)$

Paul Romero
Mar 9, 2021

Let's N to be the number of ways we can get positive integer solutions for $x^{2} - y^{2} = (x + y)\times(x - y) = 2016$ , n to be the number of divisors of 2016 and $n_{o}$ the number of odd divisors of 2016, so since $2016 = 2^{5}\times3^{2}\times7^{1} \Rightarrow n = (5 + 1)\times(2 + 1)\times(1 + 1) = 36$ .

Now we have to disregard the odd divisors $d_{oi}$ and their complements $c_{oi}$ such that $d_{oi}\times c_{oi} = 2016$ , since both (x - y) and (x + y) must be even. The odd divisors of 2016 are 1, 3, 9, 7, 21 and 63, so $n_{o} = 6$ . Therefore, $N = \frac{n - 2n_{o}}{2} = \frac{36 - 2\times6}{2} = \boxed{12}$

Optionnal differences of squares.

4 solutions

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