Testing pythagorean triples: $333,444,555$ In this page there is a necessary and sufficient test given for testing Pythagorean triples:

A simpler, more powerful test is, (by naming the even leg a): $(c − a)$ and $\large\frac{(c − b)}{2}$ are both perfect squares. This is both
  necessary and sufficient for the triple to be a PT.

Using this here,we can write $a = 444,b=333,c=555$,which means $111$ and $\frac{222}{2}=111$ must be perfect squares but it is not.Hence,that will not work. 
Is there any necessary and sufficient condition that will work for every and any (other than summing up the squares and checking for perfect squares?
NOTE: $333,444,555$ are Pythagorean triples as $3\times111,4\times111,5\times111$ for $3,4,5$ is a Pythagorean triple.
 A: Yes, there is a mistake in the phrasing of the condition. 
Every Pythagorean triple is of the form 
$$ a = 2k mn \qquad b = k(m^2 - n^2) \qquad c = k(m^2 + n^2) $$
which means that 
$$ c - a = k(m^2 + n^2 - 2mn) = k (m-n)^2 \qquad \frac{c-b}{2} = k n^2 $$
So the correct statement is that 

Let $d$ be the greatest common divisor of $c-a$ and $(c-b)/2$. Then a necessary and sufficient condition for $(a,b,c)$ to be a Pythagorean triple is that $(c-a)/d$ and $(c-b)/(2d)$ are both perfect squares. 

A: The page is not correct in stating that this relation is necessary for a Pythagorean triple, as you have shown.  For a non-primitive triple $c-a$ and $\frac{c-b}{2}$ can share a factor, $111$ in your case.  The relation is sufficient, but it will yield some non-primitive triples, such as $27,36,45$, where $c-a=9, \frac{c-b}{2}=9$.  If $c-a$ and $\frac{c-b}{2}$ do not share a factor and are not squares, it is not a PT.
A: There is more simpler way to solve (333, 444, 555) for being a triple. It obviously is a multiple of primitive triple, so extract out its common factor: 111(3,4,5); forget about common factor, prove (3,4,5) is a primitive triple; let ( 3, 4, 5 ) be a triple representing in that order (x,y,z); k=z+y = 5+4 = 9; j= z-y= 5-4= 1; j*k =x^2. So product of 9 and 1 is 9 = 3^2.So x =3. That is evidently the case. So (333, 444, 555) is a Pythagorean triple.
A: Let 333, 444, and 555 are x,y,z of a Pythagorean triple to be tested.
Rule to remember: "a given triple is Pythagorean iff there are two integers (j,k; k>j) such that square root of their product is an integer (obviously x) where k = z+y and j=z-y"
In this example: k=555+444=999, j=555-444=111; j*k = 110889; (jk)^(0.5) = 333, which is an integer and hence x.
Test is proven.
