From problem 3 here is what I have tried: a^2+b^2=c^2 => b^2=c^2-a^2 so b 2 =(c+

Question

From problem 3 here is what I have tried:

a^2+b^2=c^2

=> b^2=c^2-a^2

so

b2=(c+a)(c-a)

And since b2 is even, a,c must both be even or both be odd.

How do I then show that b^2 is a multiple of 16?

Problem 3. Let (a, b, c) be a primitive Pythagorean triple where b is even. Show that b is divisible by 4. Prove this directly without using the theorem that describes all primitive Pythagorean triples. Problem 4. Let (a, b,c) be a Pythagorean triple. Show that at least one of a and b is divisible by 3. Use this result and the result of the previous problem to prove that the area of an integer right triangle is an integer divisible by 6. Do not use the theorem that describes all primitive Pythagorean triples in this problem.

Explanation / Answer

since (a,b,c) are primitive Pythagorean triple,

all of them should be an integer and they should'nt have common divisor.

=> if b is even then a & c has to be odd.

and so,

a2+b2=c2 => b2=c2-a2=(a+c)(a-c) since Odd +odd =even, odd-odd=even      where

now since b has to be a perfect square,

=>a+c=(k12)*g (k1 is an integer) since a+c is even k1 has to be even =>k12=4m (m is an integer)

=>c-a=(k22)/g (k2 is an integer) since a-c is even k2 has to be even => k22=4n (n is an integer)

where g is any integer.

we can also say that

k12=2h/g..........h (is an integer)

k22=2t*g...........t (is an integer)

here h/g and t*g should be divisible by to for k12and k22 to be perfect even squares.

we can have 2 cases here g is even and g is odd:-

if g is even (say q2y) then h has to be atleast divisible by 2y+1, where (q is odd,y is integer >=1)

then (a+c)(c-a)=4*2y+1*t*r        (r=h/2y+1,and an integer)

                         =divisible 16 since y>=1

if g is odd (say q) then h is atleast divisible by 2,and t is also atleast divisible by 2

there fore again we get

=> b2=(4m)(4n)=16mn (where ofcourse m is h/2 & n is t/2 ,both are integers)

Also from above statements try to convince yourself that mn is a perfect square.

=>b =4l (where l=(mn)1/2)

Therefore proved

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