Pythagorean Triplets

WLOG let $b=a+1$ . $a^2+(a+1)^2=2a^2+2a+1=c^2$ $a=\dfrac {-2\pm \sqrt{4-8 (1-c^2)}}{4}$ $= \dfrac {-1\pm \sqrt{2c^2-1}}{2}$

For a to be an integer, we need $2c^2-1=n^2$ for some integer n.

$\Rightarrow n^2-2c^2=-1$

This is a negative Pell's Equation which has fundamental solution $(n,c)=(1,1)$ . Other solutions can be found by the recurrence relation $n_{i+1}=3n_i+4c_i$ and $c_{i+1}=2n_i+3c_i$ .

Hence the solutions are $(n,c)=(1,1), (7,5), (41,29), (239,\boxed{169})$

Claim: If $a, c$ are such that $a^2 + (a+1)^2 = c^2$ , then the same is true for $(a',c') = (3a + 2c + 1,\ 4a + 3c + 2).$ Proof: $(a')^2 + (a'+1)^2 = (3a+2c+1)^2 + (3a+2c+2)^2 = 18a^2 + 8c^2 + 24ac + 18a + 12c + 5; \\ (c')^2 = (4a + 3c + 2)^2 = 16a^2 + 9c^2 + 24ac + 16a + 12c + 4.$ The difference between these two expressions is $(a')^2 + (a'+1)^2 - (c')^2 = 2a^2 - c^2 + 2a + 1 = a^2 + (a+1)^2 - c^2 = 0.$ Therefore $(a')^2 + (a'+1)^2 = (c')^2$ , Q.E.D.

We know that $3^2 + 4^2 = 5^2$ , so $(a,c) = (3, 5)$ is a valid starting point. It follows that $(a',c') = (3\cdot 3 + 2\cdot 5 + 1,\ 4\cdot 3 + 3\cdot 5 + 2) = (20, 29)$ is another solution, as was given; next, $(a'',c'') = (3\cdot 20 + 2\cdot 29 + 1,\ 4\cdot 20 + 3\cdot 29 + 2) = (119, 169)$ is a solution. Indeed, $\boxed{169}$ is the answer to the problem.

Note 1: I have not proven that these are the only solutions... This method does not rule out the existence of $(20, 29) < (a^\star, c^\star) < (119, 169)$ that satisfy the equation. Thus my solution is lacking; yet I post it here because it shows a nice elementary approach to this kind of problem.

Note 2: The question is, of course, how to get the recursion relation of the claim stated above. My inspiration was the repeated fraction approximation of $\sqrt 2$ .

First, from $a^2 + (a+1)^2 = c^2 \\ 2(a+\tfrac12)^2 = c^2 + \tfrac14 \\ (2a+1)^2 = 2 c^2 + \tfrac12$ we conclude $2a + 1 \approx \sqrt 2 c. \\ \frac{2a + 1} c \approx \sqrt 2.$ Next, $\sqrt 2 = 1 + \frac 1 {2 + \frac 1 {2 + \frac 1 \dots } }.$ Thus is one approximation of $\sqrt 2 = p/q$ , the next approximation is $1 + \frac 1 {1 + p/q} = 1 + \frac q {p + q} = \frac {p + 2q} {p+q},$ i.e. $p' = p + 2q$ and $q' = p + q$ .

If $q = c$ and $p = 2a+1$ , this becomes $a' = a + c$ and $c' = 2a + c + 1$ . Thus I tried $(a, c) = (1, 2):\ \ p/q = \tfrac 3 2,\ \ a^2 + (a+1)^2 = 1^2 + 2^2 \not= 2^2; \\ (a',c') = (3, 5):\ \ p/q = \tfrac 7 5,\ \ a^2 + (a+1)^2 = 3^2 + 4^2 = 5^2; \\ (a'',c'') = (8, 12):\ \ p/q = \tfrac{17}{12}, \ \ a^2 + (a+1)^2 = 8^2 + 9^2 \not= 12^2; \\ (a''',c''') = (20, 29):\ \ p/q = \tfrac{49}{29}, \ \ a^2 + (a+1)^2 = 20^2 + 21^2 = 29^2.$ It appears that every other step results in a good equation, while in the other steps the LHS is one greater than the RHS. Thus I adjust my recursion formula by skipping a step: $(a'',c'') = (a' + c', 2a' + c' + 1) = (3a + 2c + 1, 4a + 3c + 2),$ and starting with the first valid pair $(a, c) = (3, 5)$ (or perhaps $(a,c) = (0, 1)$ ), I arrive at the claim above. The proof by induction is now straightforward.