Download Problem 3: We call a right triangle Pythagorean if each of its side

Problem 3:
We call a right triangle Pythagorean if each of its side lengths is an integer. Find all possible
values of the perimeter of a Pythagorean triangle whose hypotenuse has length 2017.
Let’s call the legs x and y. We get the following equation: 𝑥 2 + 𝑦 2 = 20172 . Subtracting x from
both sides and factoring, we get that 𝑦 2 = (2017 − 𝑥)(2017 + 𝑥) by the difference of squares.
Let’s see… we know that the difference is 2x, which means that the difference is even. We can
use this to our advantage. Let’s find squares with units digits that are 0, 2, 4, 6, and 8 apart.
First, let’s inspect the units digits of squares:
Units
1
digit of
original
Units
1
digit of
square
2
3
4
5
6
7
8
9
4
9
6
5
6
9
4
1
Let’s find squares with units digits that are 0 apart… those are every pair of the original units
digits that add up to 10 (for example 1+9=10 and they both have squares that have a units digit
of 1), as well as numbers having the same units digit.
For units digits that are 2 apart, we have units digits that end with 4 and 6, but that is it.
For units digits that are 4 apart, we have units digits that end with 1 and 5 or 5 and 9. We also
have a pair that ends with 4 and 0.
For units digits that are 6 apart, we have squares of units digits that end with 6 and 0.
Finally, for 8 apart, we have squares of units digits that end with 1 and 9.
Now, let’s experiment with the units digit of 2017-x and 2017+x. Since we want perfect square
factors, let’s see if we can get lucky by saying that both 2017-x and 2017+x are perfect squares:
X=0
1
2017-x 8
(7)
2017+x 6
(7)
2
9
3
0
4
1
5
2
6
3
7
4
8
5
9
6
5
4
3
2
1
0
9
8
I have highlighted the units digits that satisfy our units digit restrictions for squares above the
table.
Now, let’s look at a list of squares around 2017 to see if we can find any easy solutions. Based
on the table above, we only have to look at perfect squares that have a units digit of 0, 4, 5, and
9. Also note that the square roots of these perfect squares can have units digits of 0, 2, 8, 5, 3,
and 7 respectively for their units digits.
30^2= 900
40^2= 1600
50^2= 2500
32^2=1024
42^2=1764
52^2=2704
33^2=1089
43^2=1849
53^2=2809
35^2=1225
45^2=2025
55^2=3025
37^2=1369
47^2=2209
57^2=3249
38^2=1444
48^2=2304
58^2=3364
Now, we can calculate the absolute difference with 2017 and see what matches up:
30^2= 1117
40^2= 417
50^2= 483
32^2=993
42^2=253
52^2=687
33^2=928
43^2=168
53^2=792
35^2=792
45^2=8
55^2=1008
37^2=648
47^2=192
57^2=1232
38^2=573
48^2=287
58^2=1374
Ding ding ding! 35^2 and 53^2 match up! This means that x in our equation is 792. Plugging it
back in:
𝑦 2 = (2017 − 792)(2017 + 792) → 𝑦 2 = (1225)(2809)
Not letting our tables go to waste, we know that 1225 is already 35 squared and 2809 is already
53 squared. Therefore, y is equal to 35 × 53 = 1855. This means that our perimeter of our
triangle is 792+1855+2017=4664.
We know that this is the only possible answer because as we go up or down in squares, our
values will always change a slight bit since we have different differences. We can show this
using the difference of squares. Take our solution, 53^2 and 35^2 for example. By difference of
squares, if we keep going up and down, we will get different values. Going 1 up and down, we
get (54 − 53)(54 + 53)and (35 − 34)(35 + 34) which are different differences. However,
eventually we will get to a solution that is (0, 2017). This is not valid because a triangle cannot
have a 0 side length.
Therefore our only answer is 4664.