Given positive integers a, b, and c which satisfy a2+b2=c2 (a Pythagorean triple), the product abc is divisible by 60.
Note that a Pythagorean triple (a,b,c) with gcd(a,b,c)=1 (no common factors) is called a fundamental Pythagorean triple. All Pythagorean triples are either a fundamental triple or a multiple of a fundamental triple, so proving the above for fundamental triples is sufficient to prove it for all triples.
Now, let us consider a fundamental Pythagorean triple modulo 2. The integers cannot all be even, because then there would be a common factor greater than 1. If both legs are even, then
requiring that c be even as well, which contradicts the above.
Now, if both legs are odd, then c must be even:
and thus c2 must be divisible by 4. But if the legs are odd, then there are non-negative integers k and m such that a=2k+1 and b=2m+1, so
, which is not divisible by four; if both legs are odd, the sum of their squares cannot be a perfect square.
Thus, for a fundamental triple, we have one even leg and one odd leg, and an odd hypoteneuse:
Next, let us consider the triple modulo 3. There are three possible values for an integer mod 3: 0, 1, or 2≡-1. Thus, the square of an integer is congruent to either 0 or 1 modulo 3.
If both legs are divisible by three, then the hypotenuse is divisible by three as well:
and our triple is not fundamental. If neither leg is divisible by 3, then
but the square of an integer cannot be congruent to 2 modulo 3, and so the above cannot work either. Thus, one, and only one, leg of a fundamental Pythagorean triple is divisible by 3.
Now, let us consider modulo 4. There are four possible values for an integer mod 4: 0, 1, 2, or 3.
Now, we know one leg is even, and one leg is odd. Let us call by b the even leg. Then
. If a and c are both congruent to 1 modulo 4 or both congruent to 3 modulo 4, then (c–a) is a multiple of 4, and since both a and c are odd, (c+a) is even, and so is then a multiple of 8. And if one of a and c is remainder 1 and the other remainder 3, then (c+a) is divisible by four, and (c–a) is even, and so we see that in all cases, is divisible by 8. Now, suppose that b is not a multiple of four. Then b≡2 (mod 4), and so b=2k, where k is an odd integer. But then , and since k is odd, k2 is odd, and so b2 cannot then be divisible by 8. Thus, the even leg of a fundamental Pythagorean triple is divisible by 4.
Now, consider the numbers modulo 5. Considering the values of an integer and its square modulo 5, we have
|n mod 5||n2 mod 5|
To be fundamental, we cannot have both legs divisible by five. Now, if one leg is divisible by five, say a, then
and there is no contradiction. But suppose neither leg is divisible by five. If both legs have squares that give remainder 1 modulo 5, then
but from the table, we see that a perfect square cannot have that as its remainder when divided by 5. Similarly, if both legs have squares congruent to 4 modulo 5, then
which also cannot be the remainder of a square number when divided by 5. So, we are left with the remaining case of one leg having square congruent to 1 and the other having square congruent to 4. Then
, and we have c divisible by 5. Thus, one of the three integers a, b, c is divisible by 5.
Now, we’ve shown that for a fundamental Pythagorean triple:
-one of the legs is a multiple of 3;
-one of the legs is a multiple of 4;
-one of the sides is a multiple of 5.
Thus, the product abc of all three integers must be divisible by 3*4*5=60. And so abc is divisible by 60 for all Pythagorean triples (a,b,c).