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I can see that it obviously will be satisfied by the pythagorean theorem, that is:

$$3^2+4^2=5^2$$

But I am sure this isn't the way to prove the statement since you are making the assumption that it is a right triangle. I know I also can't use any of the trig functions like sin, cos, tan. Then given the sides of the triangle, can I prove this to be a right triangle?

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    See [here](http://matheducators.stackexchange.com/questions/9847/given-a-3-4-5-triangle-how-do-you-know-that-it-is-a-right-triangle). – 2017-01-12
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    There is something called *converse of Pythagoras theorem*. – 2017-01-12
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    @user3000482. You can use Cosine rule. – 2017-01-12
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    [Relevant](https://jpmccarthymaths.files.wordpress.com/2013/10/pythagoras-converse0001.pdf) – 2017-01-12

2 Answers 2

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If a triangle has two perpendicular sides with lengths $3$ and $4$, the length of the remaining side is $5$ by the Pythagorean theorem. Assume that a non-right triangle $T$ with side lengths $3,4,5$ exists. By the $SSS$ criterion of congruence, it is possible to overlap such triangle with the previous right triangle, contradiction.

Unwrapped version: by $SSS$, there is a unique triangle with side lenghts $3,4,5$, up to isometries. Since there is a right triangle with such side lengths, every triangle with side lenghts $3,4,5$ is a right triangle.

Alternative, creative version: by Heron's formula, the area of a triangle with side lenghts $3,4,5$ is $6$. That implies the orthogonality of the sides with lenghts $3$ and $4$, since $6=\frac{3\cdot 4}{2}$.

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    This answer is good, but wrapping it up in a proof by contradiction is pretty unnecessary. – 2017-01-12
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    But is there any way to prove this without using Pythagorean theorem at all? We can use pythagorean theorem when we know the triangle is right triangle, don't we? I want to prove this in as elementary way as possible. – 2017-01-12
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    @user3000482: the point is that if $BCA$ is a right triangle, $a^2+b^2=c^2$ **and vice-versa**. At some point, by contradiction or not, you have to use the Pythagorean theorem (or an equivalent version of it like Euclid's theorem), since otherwise the lengths $3,4,5$ are just unrelated. – 2017-01-12
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    @MeesdeVries: you are right, I added an *unwrapped* version of the same argument, avoiding any contradiction. – 2017-01-12
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You can either use the converse of the theorem of Pythagoras or, more elegant, the cosine law

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    I do not understand the downvote. Really. So, (+1) back. – 2017-01-12
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    @JackD'Aurizio, I'm not the downvoter, but the cosine law is not relevant as the user specifically says they cannot use trigonometric functions, and answering "the converse of Pythagoras' theorem" without reference or proof (or explanation of what it is) is basically just repeating the question back to the asker. – 2017-01-12
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    @MeesdeVries: it is not uncommon to state the cosine theorem as a generalization of the Pythagorean theorem, without making an explicit mention of the $\cos$ function, but using it as a ratio of lengths. Peter's approach makes perfectly sense along these lines. – 2017-01-12
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    My bad, I was unfamiliar with that formulation (or, at least, that that formulation carried the same name). – 2017-01-12
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    I just want to note that this problem was given in high school geometry class. Using the converse of the theorem of Pythgoras would be too much... – 2017-01-12