In number Theory Show that 60 divides the product of the len

In number Theory:

Show that 60 divides the product of the lenghts of the sides of a pythagorean triangle

Solution

Let, a,b,c be the pythagorean triple.

a^2+b^2=c^2

Case 1. Divisiblity by 3

Let none of a or b be divisible by 3.

But squares mod 3 always give residue 1

So, a^1+b^2=1+1=2 mod 3

This is not possible as:

c^2=1 mod 3

Hence one of a or b must be divisible by 3

Case 2: Divisibility by 4

We prove that none of a,b,c are divisible by 4

Assume none of a,b,c are divisible by 4.

All cannot be odd. Else, a^2,b^2 are odd and a^2+b^2 is even so c is even.

So at least one of them must be even ie of the form: 4n+2 or 4n+4

Let ,a ,b be odd and c even

So, a^2=4k+1,b^2=4m+1

a^2+b^2=4(k+m)+2=2 mod 4

which is not possible as:c=2 mod 4 so c^2=4 mod 4

So, one of a or b must be even

Let, a be even and b and c be odd

a^2=c^2-b^2

c=2k+1,b=2m+1

a^2=(c-b)(c+b)=(2k-2m)(2k+2m+2)=4(k-m)(k+m+1)

So, a=2p for some integer p

p^2=(k-m)(k+m+1)

If k , m are both even or both odd. Then k-m is even =2r

p^2=2r(k+m+1)

Hence, 2|p^2 ie 2|p

Hence, a is a multiple of 4

If k and m are of opposite parity them:k+m+1 is even and again we conclude: a is a multiple of 4

Case 3: Divisibility by 5

We show one of a,b,c is a multple of 5

Assume none of them is a multiple of 5.

Residues modulo 5 are:0,1,-1

So possible cases are:

a^2=1,b^2=-1 mod 5

a^2=-b^2=1 mod 5. Hence, a^2+b^2=0=c^2 mod 5 which is not possible.

a^2=1,b^2=1 mod 5

a^2+b^2=2=c^2 mod 5 which is not possible.

These are only possible cases where a ,band c are not divisible by 5

So one of a,b,c must be divisible by 5

Hence, the product : abc is divisible by:3*4*5=60

Hence proved.