Question

Suppose that A is a triangle with integer sides and integer area. Prove that the semiperimeter of A cannot be a prime number.

(Hint: Suppose that a natrual number x is a perfect square, and suppose that p is a prime number that divides x. Explain why it must be the case that p divides x an even number of times)

Answer 1

No, none of the lengths can be prime.

Let me write d,e,f

for ha,hb,hc, then twice the area is

2Area=ad=be=cf

Furthermore let c=x+y with x,y possibly signed lengths satisfying x2+f2=a2 and y2+f2=b2, so x,y

are (possibly signed) integers.

Consider the case in which a side length is prime, wlog let it be c

.

Then since c>d

, c∤d, but c∣ad so we must have a=tc, and similarly c>e so b=uc for integers t,u. Since a,b,c make a triangle a<b+c and b<a+c we have t<u+1,u<t+1⟹t=u and the triangle must be isosceles with a=b

.

Then x=y=c/2

, but a2=f2+x2 requires x to be an integer, so c is even and can only be prime if c=2. But then a2=f2+1 which is impossible for positive integers a,f

, so there cannot be such a triangle with a prime side.

Consider the case in which a height is prime, wlog let it be f

.

If f∣a

then f∣x and (x/f)2+1=(a/f)2 which is impossible for nonzero integers x/f,a/f. So f∤a and f∣ad⟹d=tf. Similarly f∤b and so e=uf. Then from ad=be=cf we have

c=ta=ub. Wlog a≥b, then from c<a+b we have (t−1)a<b⟹t=1 and the triangle must be isosceles with a=c

.

Then with c=ub,e=uf

and by symmetry the altitude bisecting b

(b/2)2+e2=c2(b/2)2+u2f2=u2b2f2=(b2u)2(4u2−1)

Since f is an integer and gcd(2u,4u2−1)=1 we must have 2u∣b. Since we have already shown f∤b and by assumption f is prime, then gcd(f,b)=1 and we must have b/2u=1. Then f2=4u2−1, but this is impossible for positive integers u,f, so there cannot be such a triangle with prime height.