Primitive Leg Differences

Sledgehammer to crack a walnut, but the tree of Pythagorean triples gives a quick method for this.

Per the linked article, any primitive Pythagorean triple $(a,b,c)$ has three "children" that are also primitive Pythagorean triples, formed by multiplying the vector $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ by, respectively,

$\begin{pmatrix} 1& -2 & 2\\ 2 & -1 & 2\\ 2& -2 & 3 \end{pmatrix}, \begin{pmatrix} 1& 2&2 \\ 2& 1& 2\\ 2& 2& 3 \end{pmatrix}, \begin{pmatrix} -1 & 2 & 2\\ -2& 1 &2 \\ -2& 2 & 3 \end{pmatrix}$

The key fact is that all primitive Pythagorean triples can be formed this way by starting at $(3,4,5)$ .

If the new triple is $(a',b',c')$ , then we have either $|a'-b'|=|a-b|$

or $|a'-b'|=a+b$

Starting from the triple $(3,4,5)$ , then, it's clear that the only differences we can obtain that are not more than $10$ are $1$ and $7$ , for a total of $\boxed8$ .

Solution: 8

$k = 1$ is possible because $(3, 4, 5)$ is a primitive Pythagorean triple, and $4 - 3 = 1$ .

An even $k$ is not possible:

If $x$ is even, then $x^2 \equiv 0 \pmod {4}$ , and if $x$ is odd, then $x^2 \equiv 1 \pmod {4}$ . Therefore, all squares are either $x^2 \equiv 0, 1 \pmod {4}$ .

If $a$ is even, then $b = a + k$ is even, which means $c^2 = a^2 + b^2$ is even, which means $c$ is even. But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $2$ .

If $a$ is odd, then $b = a + k$ is odd, which means $c^2 = a^2 + b^2 \equiv 2 \pmod {4}$ , which is not possible for a square.

Since $a$ cannot be even or odd, $k$ cannot be even.

$k = 3$ is not possible:

If $x \equiv 0 \pmod {3}$ , then $x^2 \equiv 0 \pmod {3}$ , and if $x \equiv 1 \pmod {3}$ , then $x^2 \equiv 1 \pmod {3}$ , and if $x \equiv 2 \pmod {3}$ , then $x^2 \equiv 1 \pmod {3}$ . Therefore, all squares are either $x^2 \equiv 0, 1 \pmod {3}$ .

If $a \equiv 0 \pmod {3}$ , then $b = a + 3 \equiv 0 \pmod {3}$ , which means $c^2 = a^2 + b^2 \equiv 0 \pmod {3}$ , which means $c \equiv 0 \pmod {3}$ . But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $3$ .

If $a \equiv 1 \pmod {3}$ , then $b = a + 3 \equiv 1 \pmod {3}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {3}$ , which is not possible for a square.

If $a \equiv 2 \pmod {3}$ , then $b = a + 3 \equiv 2 \pmod {3}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {3}$ , which is not possible for a square.

Since it is not possible for $a \equiv 0, 1, 2 \pmod {3}$ , $k = 3$ is not possible.

$k = 5$ is not possible:

If $x \equiv 0 \pmod {5}$ , then $x^2 \equiv 0 \pmod {5}$ , and if $x \equiv 1 \pmod {5}$ , then $x^2 \equiv 1 \pmod {5}$ , if $x \equiv 2 \pmod {5}$ , then $x^2 \equiv 4 \pmod {5}$ , if $x \equiv 3 \pmod {5}$ , then $x^2 \equiv 4 \pmod {5}$ , and if $x \equiv 4 \pmod {5}$ , then $x^2 \equiv 1 \pmod {5}$ . Therefore, all squares are either $x^2 \equiv 0, 1, 4 \pmod {5}$ .

If $a \equiv 0 \pmod {5}$ , then $b = a + 5 \equiv 0 \pmod {5}$ , which means $c^2 = a^2 + b^2 \equiv 0 \pmod {5}$ , which means $c \equiv 0 \pmod {5}$ . But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $5$ .

If $a \equiv 1 \pmod {5}$ , then $b = a + 5 \equiv 1 \pmod {5}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {5}$ , which is not possible for a square.

If $a \equiv 2 \pmod {5}$ , then $b = a + 5 \equiv 2 \pmod {5}$ , which means $c^2 = a^2 + b^2 \equiv 3 \pmod {5}$ , which is not possible for a square.

If $a \equiv 3 \pmod {5}$ , then $b = a + 5 \equiv 3 \pmod {5}$ , which means $c^2 = a^2 + b^2 \equiv 3 \pmod {5}$ , which is not possible for a square.

If $a \equiv 4 \pmod {5}$ , then $b = a + 5 \equiv 4 \pmod {5}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {5}$ , which is not possible for a square.

Since it is not possible for $a \equiv 0, 1, 2, 3, 4 \pmod {5}$ , $k = 5$ is not possible.

$k = 7$ is possible because $(5, 12, 13)$ is a primitive Pythagorean triple, and $12 - 5 = 7$ .

$k = 9$ is not possible:

If $x \equiv 0 \pmod {9}$ , then $x^2 \equiv 0 \pmod {9}$ , and if $x \equiv 1 \pmod {9}$ , then $x^2 \equiv 1 \pmod {9}$ , if $x \equiv 2 \pmod {9}$ , then $x^2 \equiv 4 \pmod {9}$ , if $x \equiv 3 \pmod {9}$ , then $x^2 \equiv 0 \pmod {9}$ , if $x \equiv 4 \pmod {9}$ , then $x^2 \equiv 7 \pmod {9}$ , if $x \equiv 5 \pmod {9}$ , then $x^2 \equiv 7 \pmod {9}$ , if $x \equiv 6 \pmod {9}$ , then $x^2 \equiv 0 \pmod {9}$ , if $x \equiv 7 \pmod {9}$ , then $x^2 \equiv 4 \pmod {9}$ , and if $x \equiv 8 \pmod {9}$ , then $x^2 \equiv 1 \pmod {9}$ . Therefore, all squares are either $x^2 \equiv 0, 1, 4, 7 \pmod {9}$ .

If $a \equiv 0 \pmod {9}$ , then $b = a + 9 \equiv 0 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 0 \pmod {9}$ , which means $c \equiv 0 \pmod {9}$ . But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $9$ .

If $a \equiv 1 \pmod {9}$ , then $b = a + 9 \equiv 1 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {9}$ , which is not possible for a square.

If $a \equiv 2 \pmod {9}$ , then $b = a + 9 \equiv 2 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 8 \pmod {9}$ , which is not possible for a square.

If $a \equiv 3 \pmod {9}$ , then $b = a + 9 \equiv 3 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 0 \pmod {9}$ . But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $3$ .

If $a \equiv 4 \pmod {9}$ , then $b = a + 9 \equiv 4 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 5 \pmod {9}$ , which is not possible for a square.

If $a \equiv 5 \pmod {9}$ , then $b = a + 9 \equiv 5 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 5 \pmod {9}$ , which is not possible for a square.

If $a \equiv 6 \pmod {9}$ , then $b = a + 9 \equiv 6 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 0 \pmod {9}$ . But then $(a, b, c)$ is not a primitive Pythagorean triple because $a$ , $b$ , and $c$ are all divisible by $3$ .

If $a \equiv 7 \pmod {9}$ , then $b = a + 9 \equiv 7 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 8 \pmod {9}$ , which is not possible for a square.

If $a \equiv 8 \pmod {9}$ , then $b = a + 9 \equiv 8 \pmod {9}$ , which means $c^2 = a^2 + b^2 \equiv 2 \pmod {9}$ , which is not possible for a square.

Since it is not possible for $a \equiv 0, 1, 2, 3, 4, 5, 6, 7, 8 \pmod {9}$ , $k = 9$ is not possible.

In summary, only $k = 1$ and $k = 7$ are possible, so the sum is $1 + 7 = \boxed{8}$ .

We can generate every primitive Pythagorean triplet by this identity, (4T)^2+(8T+1)=(4T+1)^2,for a^2+b^2=c^2 for c-a=1.(T denotes any triangular number).Then we take T =1 becomes 16+9=25 or 4-3=1 is one sol. Again let T=3 becomes 144+25=169 or 12-5=7.Hence we have 1+7=8.

[Observation based solution, Not a proof solution]

The only primitive Pythagorean triples with the difference between legs in the range $0 < k \leq 10$ I found were, { $3,4,5 \implies k = 1$ } and { $5,12,13 \implies k = 7$ }. There were two other triples { $8, 15, 17$ } and { $20, 21, 29$ }, but they had the same values of $k$ . Since the value of k increases, $k\nleq 10$

$\therefore \text{Sum of values of } k = 1 + 7 \implies \boxed{8}$