Question

If the length of one leg of a right triangle is 16 and the length of the other leg is odd, what are the lengths of all the sides of the triangle?

Answer 1

thats implied by 'odd' :)

Answer 2

ill brute force it , but theres probably an elegant solution

Answer 3

the sides are 16, 63 and

Answer 4

65

Answer 5

brute force isnt bad i guess as the numbers are small.. yes its giving 63, 65 haha

Answer 6

it was asked by another user sometime ago..i was stuck at finding the solution because i didnt try brute force but yeah im also pretty sure ther exists some analytic solution

Answer 7

There is a general formula for integer solutions to a^2 + b^2 = c^2

Answer 8

the 'primitive' pythagorean triples, a,b are not both even (you can factor out an even)

Answer 9

I see... a, b must have opposite parity. even-odd or odd-even

Answer 10

(when a,b,c are primitive)

Answer 11

i can look in my book, theres a chapter on it, somewhere

Answer 12

no its okay.. im just messing wid this problem as im free right now and this looks interesting..

Answer 13

my brute force was pretty lame. i used a TI 84
and did sqrt( (2x+1)^2 + 16^2), then i looked on the column for integers

Answer 14

looked on the y column, scrolled down

Answer 15

Since even + odd = odd I figured out this formula: \[\Large 2^8+(2n+1)^2=(2m+1)^2 \\ \Large 2^6 +n^2+n=m^2+m \\ \Large 2^6 = (m+n+1)(m-n) \\ \Large 2^a=(m+n+1) \text{ and } 2^b=(m-n) \\ \Large a+b = 6 \text{ and } a>b\] However I noticed each of these are off in their parity. So b=0 must be true. So adding and subtracting the two equations to solve for m and n yields \[\Large 2^6=m+n-1 \\ \Large 1=m-n \\ \Large 2^5 = m \\ \Large 2^5-1=n\] Now it's plug in time. =) \[\Large 2^6-1= 63\\ \Large 2^6+1 = 65\]

Answer 16

I found an even better solution!\[\Large 2^8+b^2=c^2 \\ \Large 2^8 = (c+b)(c-b) \\ \Large 2^x = c+b \text{ and } 2^y = c-b \text{ with } x+y=8 \\ \Large c=\frac{2^x+2^y}{2} \text{ and } b=\frac{2^x-2^y}{2} \] Now we know y=1 since b and c are both odd. \[\Large c= 2^6+1 \\ \Large b=2^6-1\]

Answer 17

Nice :)

Answer 18

This was a good question, haha got a lot of fun out of it.

Answer 19

me too i tried generating formula and missed most of above fun haha

Answer 20

Oh I'd be curious to see what you tried if you want to share.

Answer 21

a^2 + b^2 = c^2

2xy = 16 = b
x^2-y^2 = a
x^2 + y^2 = c
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from first equation we have xy=8, that implies x = 8 and y = 1
therefore a = x^2 - y^2 = 8^2 - 1 = 63
c = x^2 + y^2 = 8^2 + 1 = 65

Answer 22

that relies on identity (x^2-y^2)^2 + (2xy)^2 = (x^2+y^2)^2

Answer 23

Oh ok so is that the easiest form to remember for generating pythagorean triples? I know there exists a formula or system of equations or something to find them but I always have to look it up.

Answer 24

above formula generates all primitive pythagorean triples

i think proof shouldnt be hard..

Answer 25

we may try proof by contradiction... and i feel thats as simple/elementary as it can possibly get... yeah there are many other interesting ways to generate PTs though

Answer 26

@Kainui those are nice solutions! :)