, 2016

Solution:

There exist a triangle with side lengths $4^{2}$, $5^{2}$, $6^{2}$, which has perimeter $77$. If the sides have lengths $a^{2}$, $b^{2}$, $c^{2}$ with $0 < a < b < c$, then $a^{2} + b^{2} > c^{2}$ by the triangle inequality. Therefore $(b-1)^{2} + b^{2} \geq a^{2} + b^{2} > c^{2} \geq (b+1)^{2}$. Solving this inequality gives $b > 4$. If $b \geq 6$, then $a^{2} + b^{2} + c^{2} \geq 6^{2} + 7^{2} > 77$. If $b = 5$, then $c \geq 7$ is impossible, while $c = 6$ forces $a = 4$, which gives a perimeter of $77$.