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 6 of 6: b as the greatest integer gives no solution

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In plain words

Symmetry between " $b$ smallest" and " $b$ largest" makes this final case almost free: the same divisor bound $4$ reappears, and every candidate value of $b$ has already been dealt with, so ruling out this whole branch takes only a couple of lines.

b=\max\{a,b,c\}\ \Longrightarrow\ b\mid(b-2)^2-b^2+4b=4\ \Longrightarrow\ b\in\{1,2,4\}

Detailed analysis

Now suppose $b$ is the greatest of the three consecutive integers, so $\{a,c\}=\{b-1,b-2\}$ . The case $a=b-2$ (or symmetrically $c=b-2$ ) would require $b\mid b-2$ , which is absurd for $b>2$ . The remaining case reduces, by the same manipulation as before, to $b\mid(b-2)^2-b^2+4b=4$ , so again $b\in\{1,2,4\}$ — but $b=1$ is degenerate and $b=2,4$ were already shown in Step 5 to fail or to belong to the other case. So no triangle arises with $b$ as the largest side. Combining both cases, the only triangle satisfying the problem's conditions has sides $4,5,6$ , and it is unique.