Question

if the sides and area of the triangle have integral measures then prove that its perimeter must have an even integral measure

Answer 1

Sides of triangle ABC  be  a, b  and c  respectively.  these are integers.

semi-perimeter s =  (a+b+c) / 2        =>  perimeter  p =  2s = a+b+c
=>  to prove p is even  and hence,  s is an integer.

Area of the triangle ABC = Δ = $\sqrt{s(s-a)(s-b)(s-c)}=integer = n$
       Δ = √[ s (s - a) (s - b) (s - c) ] = integer = n
 
=> p/2  * (p/2-a) (p/2-b) (p/2-c)  =  Δ²
=> p * (p - 2a) (p - 2 b) (p - 2c)  =  16 Δ²
=>  (p² - 2(a+b) p + 4 ab) (p² - 2 pc) = 16 Δ²
=>  p⁴ -  2p³ (c+a+b)  + 4 p² [ (a+b) c + 4 ab ] - 8 abcp = 16 Δ²
=>  p⁴ - 2 p⁴  + 4 p² [ pc - c² + 4 ab ] - 8 abcp = 16 Δ²
=>  - p⁴ + 4 p³ c - 4 p² c² + 16 ab p²  - 8 abcp  =  16 Δ²
=>  p⁴  =  4 p³ c - 4 p² c²+ 16 ab p²  - 8 abcp  - 16 Δ²
           =  4 [ p³ c -  p² c² + 4 a b p²  - 2 abc p  - 4 Δ² ]

Hence, p⁴  is a multiple of 4    when   p = a+b+c,  Δ, a, b,c, are all integers.
   it is possible only if p is a multiple of 2.

Hence perimeter is an even integer.