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 3 of 6: Law of Cosines yields the key identity

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In plain words

Now the trigonometry has done its job and can be discarded: everything that matters about the angle condition has been compressed into one polynomial equation in the integers $a,b,c$ , which is exactly the kind of statement a divisibility argument can attack.

\cos\alpha=\frac{a^2+c^2-b^2}{2ac}=\frac{a}{2b} \;\Longrightarrow\; a^2c=b(a^2+c^2-b^2)

Detailed analysis

The Law of Cosines applied at vertex $B$ gives $\cos\alpha=\dfrac{a^2+c^2-b^2}{2ac}$ . Setting this equal to $\dfrac{a}{2b}$ from Step 2 and clearing denominators produces $a^2c=b(a^2+c^2-b^2)$ , an identity relating only the three side lengths.