On the regularity of the tick number

There may be a junior high school to write the last article, the exam AH ~ ~
Previously mentioned to some people, coprime hook number a,b,c (ie, a²+b²=c²) are satisfied with a law (in fact coprime satisfied, then not coprime also must meet):
When A is odd: b= (a²-1)/2
C= (a²+1)/2
When a is even: b=a²/4-1
C=a²/4+1
So, is the condition sufficient (meaning that if the a,b,c satisfies the above-mentioned rule, are these three numbers a coprime?)?
Obviously, if coprime is not required, it is absolutely satisfying and proves as follows:
When A is odd:
a²+b²=a²+[(a²-1)/2]²=a²+ (a²-1) ²/4=a²+ (a^4-2a²+1)/4=a²+ (a^4)/4-a/2+1/4= (a^4)/4+a/2+1/4= (a²+1) ²/4=[(a²+1)/2]²=c²
When a is an even number:
a²+b²=a²+ (a²/4+1) ²=a²+ (a^4)/16-a²/2+1= (a^4)/16+a²/2+1=[a²/4+1]²=c²
Next, let's discuss Coprime's question:
When A is odd: ∵a²=0 (mod a)
∴a²+1=1 (mod a)
i.e. (a²+1) and a coprime
Also because: A is odd, namely a=1 (mod 2)
∴ (a²+1)/2 also with a coprime, i.e. B and a coprime
Similarly, C and a coprime
When a is an even number: set a=2k, then:
a²/4-1= (2k) ²/4-1=4k²/4-1=k²-1
∴k² and B coprime, that is, when B is an even number (k is odd), a, a, and a maximum common factor can be 2
Similarly, A,c also have the same nature, that is, either A,b,c coprime, or a,b,c with the addition of two coprime (but in the process, many times can be converted to the former)
Then the necessity of this formula (flawed) is then proven:
When a is odd: for b,c does not satisfy the above formula, namely B≠ (a²-1)/2,c≠ (a²+1)/2:
B≠ (a²-1)/2 (mod a), c≠ (a²+1)/2 (mod a)
Also because: A is odd, namely a=1 (mod 2)
∴b≠a²-1 (mod a), c≠a²+1 (mod a)
∴b≠-1 (mod a), c≠1 (mod a)
∴b²≠1 (mod a), c²≠1 (mod a)
∴c²-b²≠0 (mod a)
And because: if A,b,c satisfies a²+b²=c², then c²-b²=a², then c²-b²=0 (mod a), contradiction, give up
When a is an even number: for b,c does not satisfy the above formula, namely B≠a²/4-1,c≠a²/4+1:
Because A is an even number, so set a=2k, then:
B≠ (2k) ²/4-1,c≠ (2k) ²/4+1
∴b≠k²-1,c≠k²+1
∴b≠-1 (mod k), c≠1 (mod k)
∴b²≠1 (mod k), c²≠1 (mod k)

∴c²-b²≠0 (mod k)
Also because: if A,b,c satisfies a²+b²=c², then c²-b²=a², then c²-b²=0 (mod a), and a=2k, so c²-b²=0 (mod k), contradiction, give up
OK, the above proof of sufficiency is purely nonsense ... A few months ago, when I wrote the Slag, now the degree of slag slightly reduced a million points, and then, falsified ... T_t ... As for the flaws in which the gods Ben have already seen through ... Inverse example ... I will say later (of course, the slag will not go to beg Qaq)
In fact, this formula can only generate some basic tick number, the following is a true, basic check the necessary and sufficient conditions of the number of shares:
Since it has taken so long to edit, I still change the letter: for x²+y²=z², the necessary and sufficient conditions are: x=2ab,y=a²-b²,z=a²+b² (A, b coprime, and A, B is a odd and even). The necessity is obvious, as to adequacy ... Eh, it's a little difficult for the slag, let's take it slow:
First of all, the definition of the problem: it is obvious that x, Y is a odd one (if all are even, then Z is even, this is not the basic tick number, if all is odd, through the 4 of the two remaining discussion, can be easily falsified), so we can set 2 | X (this symbol means that x is divided by 2 as an integer, where x is an even number), there is x=2ab.
The first is the lemma: for the indefinite equation uv=w² (w>0,u>0,u,v coprime), all its solutions can be expressed as U=a²,v=b²,w=ab (A>0,b>0,a,b coprime).
Proof: Write u,v standard decomposition =p1^e1 p2^e2 p3^e3...pk^ek,v=q1^f1 q2^f2 a3^f3......ql^fl, because U,v coprime, so p,q does not have the same prime number, therefore, W²=p1^e1 p2^e2 p3^ E3...pk^ek q1^f1 q2^f2 A3^F3......QL^FL. So E1,E2,E3......EK,F1,F2,F3......FK must be even, so u,v are completely square number, then obviously the card is finished ...
The following is a formal proof: (X/2) ²=x²/4= (z²-y²)/4=[(z+y)/2][(z-y)/2], set D to (Z+y)/2 and (Z-Y)/2 greatest common divisor, then D | ((z+y)/2+ (z-y)/2) =z,d | ([(Z+y)/2]-[(z-y)/2]) =y, and because Y,z coprime, so d=1. So (z+y)/2 with (z-y)/2 coprime. Known by lemma, Existence (Z+y)/2=a², (z-y)/2=b²,x/2=ab. That is x=2ab,y=a²-b²,z=a²+b².
The certificate is completed. The proof is quite simple and understandable ...
As for the proof of x, Y, z coprime ... Set X, Y maximum common factor to D, then D | Z= (a²+b²) and D | (a²-b²), so D | A²,d | b², but a, b coprime, so a²,b² coprime, D is equal to 1, so x, y coprime, and then can also launch X, Y, z coprime.
A counter-example ... A counter-example ... A counter-example ... Make your own A/b to generate it ... The school of Slag writing has gone ...
Some students do not understand a lot of symbols, so many things are handwritten, good trouble to say ...
Beautiful typesetting, a kind of indentation of the process of the feeling ...

On the regularity of the tick number