# Another proof for an infinite number of Pythagorean triples

I’m not sure if this has been mentioned before (and I truly apologize if someone thought about it already) , but I tried to adopt a geometrical approach for the proof for an infinite number of Pythagorean triplets. Please read it once and please point out any mistakes .

Any Pythagorean triplet can be expressed in the form of $$x^2+ y^2=r^2$$ (r being an integer). Keeping $$r$$ constant , we can easily conclude that the equation is that of a circle. Rearranging the equation gives $$y= \sqrt{r^2-x^2}$$. (considering only natural numbers) . Now the domain of $$x$$ is $$-r . Considering the base case , $$r=1$$ there is at least one integer pair that satisfies its equation , it being $$(0,1)$$. For $$r>1$$ , any integral value of $$x$$ gives a corresponding integral value of $$y$$, while the condition $$-r still holds true. As there are infinite number of natural numbers $$r$$ , the result follows.

I would really like to know if I’ve missed on something so please guide me . Thank you !

• "As there are infinite number of natural numbers r , the result follows." What you've shown is that there are an infinite number of trivial triples. This does prove your claim, but not in a satisfying way. – Don Thousand Nov 12 '18 at 16:50
• "For $r>1$, any integral value of $x$ gives [via $y=\sqrt{r^2-x^2}$] a corresponding integral value of $y$, while the condition $−r<x<r$ still holds true." Um ... Have you tried this? Take, for example, $r=7$. No non-zero integral value of $x$ between $-7$ and $7$ yields an integral value of $y$. – Blue Nov 12 '18 at 17:10
• @Blue except of course the trivial case of $x=0$. – Don Thousand Nov 12 '18 at 17:19
• @Aditi In other words, this proof just doesn't work for non - trivial tuples (trivial tuples are those with $0$ as one of the numbers). I think you should try a completely different approach. – Don Thousand Nov 12 '18 at 17:21

For example, consider the equation $$y= \sqrt {16-x^2}$$
You have claimed that for any integral value of $$x$$ you get an integral value of $$y$$
For $$x= 1, 2, 3$$ you get $$y= \sqrt {15} , \sqrt {12}, \sqrt {7}$$ and none of these numbers are integers.
The only integers that we get out of that are $$0$$ and $$4$$ which are trivial solutions to $$x^2+y^2=16$$
• Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that? – Aditi Nov 12 '18 at 17:13
• I think modifying the proof wouldn’t be very useful as it would just prove trivial cases for numbers like $r=4$ . Thank you for pointing the mistake ! – Aditi Nov 12 '18 at 17:20