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This is a problem from Project Euler, problem 94.

The problem asks about isosceles triangles with integer sides (differing by 1 unit, e.g, 5-5-6) and integer area, which are known to be Heronian Triangles.

Now as per the wiki, all heronian isosceles triangles have sides of the form:

$$ a = u^{2} + v^{2} \\ b = 2(u^2 - v^2) $$

for coprime integers u and v with u>v.

As per the question, the difference between a and b is 1, which reduces the equations to

$$ u^2 - 3v^2 = 1 \space for \space b > a \\ 3v^2 - u^2 = 1 \space for \space a > b $$

Clearly both are of the form of Pell's equation. And the second being a negative Pell equation with D = 3, it is not solvable effectively removing any possibility of triangles with a > b.

As I solved for the 1st equation, I did not get all the possible Heronian isosceles triangles. However, on the web I can find solutions considering the case of a > b as well.

Where am I doing wrong ? Isn't the question asking about Heronian isosceles triangles ? And, if the wiki is right, how can we have triangles with a > b when the corresponding Pell's equation is not solvable ?

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The formula you wrote down for generating the Heronian triangles is not quite correct. For example, we can take $u$ and $v$ odd, $a=\frac{1}{2}(u^2+v^2)$, $b=u^2-v^2$, with $u\gt v$.

That will give you the Pellian $3v^2-u^2=2$, which does have solutions.

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  • $\begingroup$ I see. Looks like I didn't notice the part where they say "All isosceles Heronian triangles are given by rational multiples of". $\endgroup$ – gaganbm Dec 14 '13 at 20:49
  • $\begingroup$ I just checked, they kind of get it right, though they should have given an iff condition, since not all rational multiples work. $\endgroup$ – André Nicolas Dec 14 '13 at 21:03

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