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 4 of 6: b must divide a²c, so b cannot be the middle integer

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In plain words

This is the crucial pruning step: instead of checking all $3!=6$ orderings of which side is $a$ , $b$ , $c$ , coprimality among consecutive integers instantly rules out half the cases, leaving only two configurations to examine by hand.

b \mid a^2c, \qquad a<b<c\ \text{or}\ c<b<a\ \Longrightarrow\ \gcd(b,a)=\gcd(b,c)=1

Detailed analysis

Since $a,b,c$ are positive integers, the identity $a^2c=b(a^2+c^2-b^2)$ shows $b\mid a^2c$ . If $b$ were strictly between $a$ and $c$ (the middle one of the three consecutive integers), then $b$ would be coprime to both its neighbors $a$ and $c$ (consecutive integers are coprime), so $b$ could not divide $a^2c$ unless $b=1$ , which is impossible in a nondegenerate triangle. Hence $b$ must be the least or the greatest of the three consecutive integers.