Question: A Canadian male has recently had a Prostate Specific Antigen (PSA) test as to determine if he has prostate cancer. The false-positive rate of a PSA test is 14%. If he does have prostate cancer, PSA test will be positive 81% of the time. 

Because this male is showing symptoms that are consistent with prostate cancer, it is assumed that the chance he has prostate cancer prior to taking the PSA test is 0.17. 

Part (a) What is the probability that the PSA test will yield a positive result? 

Part (b) If the PSA test gives a positive result, what is the probability that he does not have prostate cancer? 

Part (c) Suppose the PSA test result is negative, indicating that he does not have prostate cancer and his symptoms are a result of something else. What is the probability that he does have prostate cancer?

I've already solved A and B but for some reason I can't get C right. I have all the numbers i need in a table and am using this equation. P(D|-) = P(Dn-)/P(-) where D is having the disease and - means the result was negative.

Answer: 
If the primitive right triangle has generators $m$ and $n$, with legs $2mn$ and $m^2-n^2$ and hypotenuse $m^2+n^2$, then $m^2+n^2 = k^2$, and $(m,n)=1$ since the original triangle is primitive. This defines a new right triangle, which must have generators $p, q$:
$$m = 2pq,\quad n = p^2-q^2,\quad k = p^2+q^2.$$
Then the area of the original triangle is
$$2mn(m^2-n^2) = 2pq(p^2-q^2)(4p^2q^2-(p^2-q^2)^2.$$
Since $(p,q)=1$, we have $p^2-q^2\equiv 0\mod{2}$, so the area is divisible by 4. Also, either one of $p$ and $q$ is divisible by $3$, so the area is divisible by $6$. Finally, if neither $p$ nor $q$ is divisible by 7, it's not hard to see that $4p^2q^2-(p^2-q^2)^2$ is, using the fact that $p^2, q^2\equiv 1, 2, 4\mod{7}$.
33,56,65 and the rest are not perfect triangles. The hypotenuse is not a perfect square in any of them