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

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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 =

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.

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 =

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.