Timeline for Determine the number of integral points on hypotenuse of a right triangle
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2017 at 17:07 | history | edited | G Cab | CC BY-SA 3.0 |
amended for irrational
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| Oct 21, 2017 at 16:24 | comment | added | G Cab | @nickgard: yes, you are right, I forgot that the only exception is when $a$ and $b$ are particular combinations of the same irrational number, in which case there is just one solution. Thanks for signalling, I will amend the answer. | |
| Oct 21, 2017 at 15:51 | comment | added | nickgard | "Clearly, if $a$ or $b$ are irrational there are no solutions." I don't think this is correct. Example: I can draw a line from $(\sqrt{2},0)$ through $(1,1)$ to the $y$-axis at $(0,2+\sqrt{2})$. $$\frac{1}{\sqrt{2}}+\frac{1}{2+\sqrt{2}}=1$$ | |
| Oct 21, 2017 at 14:52 | history | edited | G Cab | CC BY-SA 3.0 |
edited central part
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| Oct 21, 2017 at 13:58 | history | answered | G Cab | CC BY-SA 3.0 |