Can a Triangle with Prime Number Sides Have a Whole Number Area?

[kaliprasad]

Prove that if the sides of a triangle are prime numbers its area cannot be whole number.

 

[RLBrown]

Prove that if the sides of a triangle are prime numbers its area cannot be whole number.

Spoiler

View attachment 3168

From both forms, the RHS is odd if no side length is 2.
If only one side is 2, the RHS still does not yield the factor of value 16 required by the LHS.

 

[kaliprasad]

Spoiler

Let the sides be $a,b, c$ and $a \le b\le c$

now there are 2 cases

$a = 2$ or all are odd
if A is area then $A^2 = \dfrac{( a+b-c)(a+b+c)(a-b+c)(b+c-a)}{4}$if all are odd then all 4 terms on the numerator of RHS are odd then $A^2$ cannot be integer so A cannot be whole number

case 2:
for $a= 2$ and $b = 2$ or $a= 2$ and $b != 2$

$a = 2\, b =2 \, => c = 2\, or\, 3$

$a =2\, b = 2\, c = 2 => A^2 = \dfrac{6*2^3}{4} = 12$ so A is not integer

$a =2\ , b= 2\, c = 3 => A^2 = \dfrac{7 * 1 * 3 * 3}{4}=\dfrac{3^2*7}{2^2}$ so A is not integerif $b\ne 2$ then $b= c$ because if $c \gt b$ then $c\ge b+2$ or $a+b\le c$

so we get $A^2 = \dfrac{(2+2b)* (2b-2)* b^2}{4}= \dfrac{b^2(b^2-1)}{4}$ cannot be a perfect squareso no solution