If you add the squares of the ages of two grandchildren, then take the square root, you have the age that their grandmother was 50 years ago. If the grandmother is less than 100 years old, what are the ages?
(Child 1 age)2 + (Child 1 age)2 = (Grandmother age − 50)2
so we have a Pythagorean triple a2 + b2 = c2 where b = a + 3
This is a problem for younger students, so probably the only Pythagorean triple they know is
32 + 42 = 52
To scale that up to get the two smaller numbers differing by 3, multiply through by 3
92 + 122 = 152
So the grandchildren are 9 and 12, and the grandmother is 65.
The next Pythagorean triple a2 + b2 = c2 with b = a + 1 (which can then multiplied through by 3 to get the two smaller numbers differing by 3) is 20, 21, 29. That would give
602 + 632 = 872
with the grandchildren 60 and 63, and the grandmother 137, but we’re told the grandmother is less than 100.
There are no primitive Pythagorean triples (that is, triples with a, b, and c coprime) with b = a + 3. Why not?
If (a, a+3, c) is a primitive Pythagorean triple, then
c2 = 2a2 +6a + 9
a ≠ 0 mod 3 because otherwise both a and c would be divisible by 3
so a = 1 or 2 mod 3, and a2 = 1 mod 3. But then c2 = 2 mod 3 which is impossible. ◼
Since 3 is a prime number, the only ways to get Pythagorean triples with the two smallest numbers differing by 3 are:
- primitive Pythagorean triples with the two smallest numbers differing by 3, none of which exist, or:
- 3 × (primitive Pythagorean triples with the two smallest numbers differing by 1)
There are infinitely many primitive Pythagorean triples whose two smallest numbers differ by 1
That follows from the fact that for n=2 there are infinitely many whole-number solutions for X and Y of the negative Pell equation
But the next primitive Pythagorean triple with the two smallest numbers differing by 1, after 3, 4, 5, is 20, 21, 29 (as above)
and next after that 119, 120, 169, giving grandchildren 357 and 360 years old…