Given positive integers *a*, *b*, and *c* which satisfy *a*^{2}+*b*^{2}=*c*^{2} (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 *c*^{2} must be divisible by 4. But if the legs are odd, then there are non-negative integers *k* and *m* such that *a*=2*k*+1 and *b*=2*m*+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*=2*k*, where *k* is an odd integer. But then , and since *k* is odd, *k*^{2} is odd, and so *b*^{2} 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 |
n^{2} 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*).

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