## Monday Math 144

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 (ca) 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 (ca) 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
0 0
1 1
2 4
3 4
4 1

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).