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 $-r<x<r$ . 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 $-r<x<r$ 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 !

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    $\begingroup$ "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. $\endgroup$ – Don Thousand Nov 12 '18 at 16:50
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    $\begingroup$ Please check: youtube.com/watch?v=QJYmyhnaaek $\endgroup$ – Mohammad Zuhair Khan Nov 12 '18 at 16:55
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    $\begingroup$ "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 $y$. $\endgroup$ – Blue Nov 12 '18 at 17:10
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    $\begingroup$ @Blue except of course the trivial case of $x=0$. $\endgroup$ – Don Thousand Nov 12 '18 at 17:19
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    $\begingroup$ @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. $\endgroup$ – Don Thousand Nov 12 '18 at 17:21

Your statement is correct but your proof is not.

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$

  • $\begingroup$ Alright , but there’s atleast one integer pair corresponding to each $r$ right ? Can I modify my proof to reflect that? $\endgroup$ – Aditi Nov 12 '18 at 17:13
  • $\begingroup$ 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 ! $\endgroup$ – Aditi Nov 12 '18 at 17:20
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    $\begingroup$ @Aditi Thanks for your comment. You are correct, your proof only gives us trivial solutions but I encourage you to keep up the good work. $\endgroup$ – Mohammad Riazi-Kermani Nov 12 '18 at 17:30

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