Problem 1

Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.

Step 5 of 6: b as the least integer forces the triangle (4,5,6)

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In plain words

This is the case that actually produces a triangle: pinning down $b$ via a divisor list of only three candidates, then eliminating two of them by hand, is a very efficient way to search an infinite family of integer triples.

b=4,\ (a,c)=(6,5) \;\Longrightarrow\; (a,b,c)=(6,4,5)

Detailed analysis

Suppose $b$ is the least of the three consecutive integers, so $\{a,c\}=\{b+1,b+2\}$ in some order. If the larger neighbor $b+2$ plays the role of $a$ , the divisibility $b\mid a^2c$ becomes $b\mid(b+2)^2$ , and since $b\mid(b+2)^2-b^2-4b=4$ , we get $b\in\{1,2,4\}$ ; checking $b=1$ (degenerate) and $b=2$ (forces $a,c=3,4$ , which fails the identity) directly, only $b=4$ survives, giving $a^2c=b(a^2+c^2-b^2)=180$ and hence $a=6$ , $c=5$ : the triangle with sides $4,5,6$ . If instead $b+2$ plays the role of $c$ , the same reasoning $b\mid b+2$ forces $b\in\{1,2\}$ , both already excluded, so this sub-case gives nothing new.