What is the maximum number of positive integers among which any three are sides of an obtuse triangle?

I can find four, $11,11,16,20$. Is it possible to get five or more? We need $a^2+b^2<c^2$ and $a+b>c$ for all $a,b,c$.

  • $\begingroup$ The smallest possible quadruple is: $6,6,9,11$. The first case where all four numbers are distinct (and hence no two of the four obtuse triangles are congruent): $7,8,11,14$ $\endgroup$ May 2, 2016 at 10:14
  • $\begingroup$ @JeppeStigNielsen: How did you get those solutions? $\endgroup$ May 2, 2016 at 11:01
  • $\begingroup$ @RoryDaulton I used brute force, a computer program. I ran through a lot of integer quadruples $(a,b,c,d)$ where (WLOG) $0<a\le b\le c\le d$. I checked a bunch of inequalities to make sure all four combinations were in fact triangles, and to check that those triangles were in fact obtuse. Upon thinking a bit more, when $0<a\le b\le c\le d$ is known, we need to check only $a+b>d$ to make sure all combinations are triangles (triangle inequality). And we need to check only $b^2+c^2<d^2$ and $a^2+b^2<c^2$ to make sure every triangle is obtuse (from law of cosines, generalizing Pythagoras). $\endgroup$ May 2, 2016 at 12:49

1 Answer 1


Suppose we have five positive integers $a_1\le a_2\le a_3\le a_4\le a_5$ such that any three are the sides of an obtuse triangle. Then by repeated substitution of $a_n^2+a_{n+1}^2<a_{n+2}^2$, $$a_5^2>a_4^2+a_3^2>2a_3^2+a_2^2>3a_2^2+2a_1^2$$ But $a_1+a_2>a_5$, so $$(a_1+a_2)^2>a_5^2>3a_2^2+2a_1^2=(a_2+a_1)^2+a_2^2+(a_2-a_1)^2$$which is impossible.

  • $\begingroup$ I guess this works for five positive real numbers as well. $\endgroup$ May 2, 2016 at 9:14

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