Define an "integer 3-relationship" as a function $f(a,b,c)$ of three integer variables, together with the condition that this function must equal zero. Two examples would be the "Pythagorean triplet relationship" with $f(a,b,c) \equiv c^2 - (a^2 + b^2) $ and the "near-miss Pythagorean triplet relationship" with $f(a,b,c) \equiv c^2 +1 - (a^2 + b^2) $.
For any given integer 3-relationship based of some $f$, there may be many triplets that satisfy the condition. For example, any triplet $$ \{k(m^2 - n^2), \,2k m n, \, k(m^2+n^2)\}\,\, (k,m,n \in \Bbb N) $$ satisfies the Pythagorean triplet relationship.
In some cases, there may be a chain of $n$ triplets $$ \{a, b, c\},\, \{b, c, d\},\, \{c, d, e\},\, \cdots $$ such that each of these triplets satisfies the same integer 3-relationship based of some $f$, and the last two entries in each triplet match the first two entries in the next. For example, with $f(a,b,c) = c^2 +1 - (a^2 + b^2)$, the triplets $\{7,\, 11,\, 13\}, \{11,\, 13,\, 17\}$ form a 2-chain because $$ 7^2 + 11^2 = 13^2+1 \\ 11^2 + 13^2 = 17^2 + 1 $$ I am concerned with looking for chains of Pythagorean triplets.
Prove that (other that trivial chains with $a$ or $b$ zero) there are no 2-chains of Pythagorean triplets (or disprove that conjecture by providing an example Pythagorean 2-chain).
I think I have a proof along the lines of starting with a "primitive" 2-chain (with no common factor in $a$ and $b$), applying the general form of Pythagorean triplets given above, and re-distributing the factors in $2 m n$ to form the $2 r s$ of the next triplet, so that $ 2mn = 2rs$ and $m^2 + n^2 = r^2 - s^2$. At that point I can solve a quadratic equation for one of the factors and the condition that the discriminant be a perfect square leads to the construction of a different 2-chain with smaller numbers -- and reduction ad absurdium applies.
But my proof is not what I consider solid...
For a chain of Pythagorean triplets we have$$b^2=c^2-a^2$$and$$b^2=d^2-c^2.$$ Hence $a^2,c^2$ and $d^2$ are in arithmetic progression with common difference $b^2.$
A proof that three squares in arithmetic progression cannot have a square as the common difference was given by Fermat and is known as Fermat's right triangle theorem. (The proof is included on the linked page.)
A number that can be a difference of three squares in AP is known as a congruum and it is known that these numbers are those that are exactly four times the area of a Pythagorean triangle. Hence, if a congruum is a square, then the area of the Pythagorean triangle must also be square.
Fermat's proof relies on showing that if there exists a Pythagorean triangle with an area that is a square number then a smaller such Pythagorean triangle must exist.