Find out if the (3,4,5) is the only possible right triangle whose semiperimeter equals its area.?

Question

Please list some other such right triangle if possible. Also, I'd really appreciate a proper proof if (3,4,5) is the only such right triangle
The sides of the triangle must be positive integers.

Answer 1

We will solve two equations:

(a+b+c)/2 == a*b/2
and
a^2+b^2==c^2

By solving for c in the first equation and substituting it into the second we get:

a^2+b^2==(a^2)*(b^2)-2*(a^2)*b-2*a*(b^2)+2*a*b+a^2+b^2

This reduces to:

(a^2)*(b^2)-2*(a^2)*b-2*a*(b^2)+2a*b == 0

Then let us factor:

a*b*(a*b-2a-2b+2) == 0

Since we are looking for positive integers a nor b can be 0, so

(a*b-2a-2b+2)==0

When we solve for a we get:

a = 2*(b-1)/(b-2)

Keep in mind this only applies when b !=2

So for a to be positive, b >= 2, when b=3, a=4 and when b=4, a=3, if b>4 we can provide a simple limit argument that a will not be an integer (since for b>2, as b goes to infinity, a monotonically decreases toward 2), so b>4 is not a possibility. This means (a,b)=(3,4) or (a,b)=(4,3) are the only possibilities.

In both cases, we see that c=5.

For b=2, (a*b-2a-2b+2)==0 yeilds a contradiction: -2==0.

Hence we see that 3, 4, 5 and 4, 3, 5 are the only possibilities.

Answer 2

Yes I believe it's the only possible solution.

x^2 + y^2=z^2 can be solved by

x= a^2-b^2 y=2ab z=a^2+b^2 and then assigning integer values to a, b and c with your condition xy= x+ y + z

=> (a^2-b^2)(2ab)=(a^2-b^2)+2ab+(a^2+b^2)

=> (a^2-b^2)(2ab)= 2a^2+2ab
(a^2-b^2)(2ab)=2a(a+b)
(a-b)(a+b)(2ab)=2a(a+b)
(a-b)b=1
ab-b^2-1=0
only a=2 and b=1 will solve this equation

we can write ab=b^2+1

Dividing by b both sides
a=b+1/b

since a,b are integers b+1/b can be integer only for b=1 => a=2
=>x=2^2-1^2=3 , y=2*2*1=4, z=2^2+1^2=5
which is the only possible solution.

Answer 3

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