Is it possible to have an even integer hypotenuse and odd integer legs (perpendicular and base) in a right triangle? If yes, please give an example. If no then please prove that.


2 Answers 2



Suppose you have $A^2+B^2=C^2$ where $A=2a+1, B=2b+1, C=2c$

Substitute, expand and then take out terms with factors of $4$


Suppose that you are right and there is a possible right triangle such that $a=2k+1, b=2m+1$ and $c=2n$ {legs are odd while hypotenuse is even} and you have $a^2+b^2=c^2$.

Now on expanding (substituting for $a, b, c$) you will get that: $$(2k+1)^2+(2m+1)^2=(2n)^2$$ $$4(k^2+m^2+k+m)+2=4n^2$$ On dividing the equation by $2$, you will get $$2(k^2+m^2+k+m)+1=2n^2$$

Notice that term on left is odd while the term on right is even. A contradiction. So, you were wrong and therefore there does not exist such a right triangle

  • 3
    $\begingroup$ I find the contradiction even more obvious if the equation is divided by 4. The left side then clearly is not an integer (given all variables are integers, the bracketed sum is one, to which we add 1/2), while the right side clearly is. I know, it's the same statement modulo 4, but still. $\endgroup$ Nov 29, 2016 at 17:32
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    $\begingroup$ I prefer dividing the equation by 2, which yields an odd integer on the left equalling an even integer on the right, violating parity. I think that's about as intuitive as equating a half-integer to an integer. $\endgroup$
    – S. G.
    Nov 29, 2016 at 17:36
  • $\begingroup$ I have taken your suggestion into account as it will improve the answer.Thanks @S.G. $\endgroup$ Nov 29, 2016 at 18:41
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    $\begingroup$ It's a bit confusing to say "you were wrong OR there does not exist such a triangle." Logically, what is true is "you were wrong AND there does not exist such a triangle"! I know that by "OR", you mean "to put it another way"; but in a mathematical context that is not so clear. A better word is THEREFORE. $\endgroup$
    – TonyK
    Nov 30, 2016 at 0:20
  • $\begingroup$ Yes @TonyK I was just saying same thing in two ways. Thanks for suggestion $\endgroup$ Nov 30, 2016 at 1:18

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