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Can you explain the notation on the right?

So 60 is a factor of xyz or a multiple?

60 is a factor of \(xyz\).

Are we assuming x,y,z are whole number cause if not that expression is definitely not always true.

The correction is up.

60 = 2^2*3*5
let m=2

it seems all multiples of
n* 3,4,5 is solution of the given condition;

n=4,
2* 4 * 15* 17 <-- multiple of 60

exper u let n=1 and i think we must work on general form

n=5,
10*24*26
n=6
.... probably yes for n>1

let's try induction.

one of m,n is even so \[4|2mn\]so\[4|abc\]we need to prove \(3|abc\) and \(5|abc\)

m * (m+1)(m-1) for any value of m, this is divisible by 3

m * (m+1)(m-1)(m^2 + 1)
^ ^ ^
can be of the form 5n+2 or 5n+3

(5n+2)^2 +1 = .... 2^2 + 1 = 5
(5n + 3)^2 + 1 = .... 3^2 + 1 = 10
both cases divisible by 5

for all n>1, the Pythagorean triplet is divisible by 60 (probably)

so \(3,4,5|abc\) it gives \(60|abc\).

o.O

what the hell is that ....

can't follow ... probably not my stuff.

@mukushla I copied it from your pdf

lol

4. extra points to whomever gets the reference

if both of \(m,n\) are odd\[m^2-n^2\equiv m^2+n^2\equiv0 \ \ \text{mod} \ 2\]

Yep. Nice job!