help using complex nos... cuz i forgot the method

[static vapor]

i'd done it.. but my sir telling me there is another way.. i am eager to know it now using complex numbers 🙂

[molten zealot]

Just take it mod x^2+2 and equate to 0.

[molten zealot]

So, the remainder is (b-2)x+(c-2a), which means b=2 and c=2a. Thus, a=1,...,10.

[tropic jay]

this is the same conclusion I came to, but I'm not quite sure why a=0 , so x³+2x is not an answer. Is 0 not counted as a natural number here? I mean 11 is not a possible answer, but I don't see why thats not it.
Also this is not using complex numbers, so not the solution the poster was searching for.

[molten zealot]

There are different math traditions for the starting value of natural numbers depending on the country. Probably they start it with 1 in India. I thought that he had a solution with complex numbers and wanted the one without. But yes, now I see that he probably wants complex numbers.

[tropic jay]

okay thank you

[tulip pulsar]

Substitute x=sqrt2i,-sqrt2i. Equate real and imaginary parts to 0. You'll get a condition ||c=2a, b=2||. Then you can get the answer with condition a,b,c<=20

[static vapor]

yeah.. thanks @tulip pulsar

[tulip pulsar]

Np. My shift btw 😅

You can mark it solved if you are done

[static vapor]

how to mark it as solved

@tulip pulsar

[strange veldt]

you can try .solved

[static vapor]

.solved