Question

All Pythagorean triples where area = perimeter (numerically)?

Answer 1

More algebra...:-)

Answer 2

5,12,13
using trial and error method ^^

Answer 3

That's one.....

Answer 4

what do you mean?

Answer 5

By what?

Answer 6

one more: 6,8,10

Answer 7

That's two....

Answer 8

hehe...

Answer 9

ok, general form tripel pythagorean number is :
a=x
b=(x^2-1)/2
c=(x^2+1)/2
with a<b<c (a,b,c are sides of a right-triangle)
perimeter = area
x + (x^2-1)/2 + (x^2+1)/2 = x(x^2-1)/2, simplify it we get :
x^3 - 4x^2 - 5x =0
x(x^2 - 4x - 5) = 0
x(x-5)(x+1)=0
x=5 satifisfy it (so, a=5, b=12, c=13)

with same way, for
a=2x
b=x^2-1
c=x^2+1
perimeter = area
2x+x^2-1+x^2+1 = 2x(x^2-1)/2, simplify it we get :
x^3 - 2x^2 - 3x = 0
x(x^2 - 2x - 3) =0
x(x-3)(x+1)=0
x=3 satifisfy it (so, a=6, b=8, c=10)

i have done for
a=kx
b=k(x^2-1)
c=k(x^2+1)
with k=3,4,5,6...
there are x satisfy it but the result just reflication of on... and its rest not satisfy for x which make for a,b,c integer number

Answer 10

so, just 2 triple only...

Answer 11

3,4,5 is a triple, too!

Answer 12

but, its area not equal its perimeter...

Answer 13

Yes, good enough:-)

Answer 14

thanks...

Answer 15

you can have an easier algebra by eliminating z from

x^2+y^2=z^2 and xy/2 = x + y +z leads to

xy -4x -4y +8 = 0 ->

(x-4)(y-4) = 8 etc.

Answer 16

just combine of (x-4)(y-4) = 8, and we will get 2 triple only are :
(x,y,z) = 5,12,13 and 6,8,10
but yours easier than me :)