2021/029) If (a,b,c) is a Pythagorean triple prove that $60 | abc$

 we have 60 = 3 * 4 * 5.

For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple

We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
we have 60 = 3 * 4 * 5.
For 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor
of a or b or c and 5 is a factor of a or b or c. We shall prove the same for
primitive Pythagorean triple
We now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
we have 60 = 3 * 4 * 5.
for 60 to be factor of abc we need to show that 3 is a factor of a or b or c. 4 is a factor of a or b or c and 5 is a factor of a or b or c. We shall prove the same for primitive Pythagorean triple
we now prove the 3 one by one and we shall be proving for primitive triples
1) For primitive Pythagorean triple one of the legs is divisible by 3
Proof:
let no leg be divisible by 3
then we have the 2 sides a and b as $a=3n\pm 1$ and $b=3m\pm 1$
so we have $c^2 = a^2 + b^2= (3n\pm 1)^2 + (3m\pm 1)^2 = (9n^2 \pm 6n + 1) + (9m^2 \pm 6m + 1)$
or $c^2 = 3(3n^2+3m^2 \pm 2n \pm 2m) + 2$
but dividing a square number by 3 does not leave a remainder 2 so this is not possibles. So atleast one side shall be divisible by 3.
2) Exactly one of a, b is divisible by 4
before proving this 1st we need to prove that exactly one of a, b is divisible by 2
because it is primitive so both cannot be divisible by 2 so let us assume that both are odd
so let a = 2x + 1 and b= 2y + 1
so $a^2+b^2 = (2x+1)^2 + (2y+1)^2 = 4x^2 + 4x + 1 + 4y^2 + 4y + 1 = 4(x^2+x+y^2+y) +2$
but this cannot be a perfect square as if it is even then it leaves a remainder 0 when divided by 4.
without loss oe generality let a be odd and b even.
Now a is odd so $a= 2x + 1$
There are 2 cases for b. either b is divisible by 4 or not 5
If b is not divisible by 4 then b is of the form $4y+2$
So $a^2 + b^2 = (2x+1)^2 + (4y+2)^2 = 4x^2+4x + 1 + 16y^2 + 16y + 4$
$= 4x(x+1) + 1 + 16y(y+1) + 4$
We have $a^2 + b^2 \equiv 5 \pmod 8$
But because if c is odd say c = 2t + 1
$c^2 = (2t+1) ^2$
Or $c^2 = 4t^2 + 4t + 1 = 4t(t+1) + 1$
So $c^2 \equiv 1 \pmod 8$
Hence $a^2 + b^2 \equiv 5 \pmod 8$ is not possible so b is divisible by 4
3) Either a or b or c is divisible by 5
If either a or b is divisible by 5 then we are through. So let us assume that neither a nor b is divisible by 5 . working in mod 5 we have
$x^2 \in \{0,1,4\} \pmod 5$
as a and b are not divisible by 5 so
$a^2 \in \{1,4\} \pmod 5$
$b^2 \in \{1,4\} \pmod 5$
if we choose $a^2=1 \pmod 5$ and $b^2=4 \pmod 5$ or $a^2=4 \pmod 5$ and $b^2=1 \pmod 5$ then only the sum could be a perfect square and and other combination does not satisfy the criteria working on mod 5 (we get 2 or 3) and this gives $c^2=0 \pmod 5$ or c is divisible by 5.
As we have proved that 3 divides a or b so 3 divides abc, 4 divides a or b so 4 divides abc and 5 divides a or b or c so 5 divides abc. hence 60 is a factor or abc.
Proved