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I have written a program that calculates the GCD of 2 polynomials using long division (Euclid's algorithm/theorem).

I am trying to find the GCD of x^2 - 3 and x + 5, which I already know is 1. However, the division process does not give me this answer. I will step through what my program does.

Firstly, it divides x^2 by x to get x. x + 5 is then multiplied by x to get x^2 + 5x. It then subtracts x^2 + 5x from x^2 - 3 to get -5x - 3. It then divides -5x by x to get -5. x + 5 is then multiplied by -5 to get -5x - 25. It then subtracts -5x - 25 from -5x - 3 to get 22. The degree of 22 (0) is less than the degree of x + 5 (1), so it stops, and passes the remainder of 22 to the program.

The program then sets up another long division, but this time divides x + 5 by 22. Firstly, it divides x by 22 to get 1/22 x. 22 is then multiplied by 1/22x to get x. It then subtracts x from x + 5 to get 5. It then divides 5 by 22 to get 5/22. 22 is then multiplied by 5/22 to get 5. It then subtracts 5 from 5 to get 0. Remainder is 0, so program stops.

According to Euclid's algorithm/theorem, the GCD would then be 22, as it was the divisor when the remainder was 0, but the GCD is in fact 1 for polynomials x^2 - 3 and x + 5. Why doesn't this work, and how would I fix it? Thankyou.

David Holden
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CoopDaddio
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Remember that you're working over real polynomials, not integer polynomials. So as far as divisibility is concerned, the polynomial $22$ is equivalent to the polynomial $1$.

That means that the Euclidean algorithm gives $$ \gcd(x^2-3, x+5)\\ =\gcd(x^2-3-x(x+5), x+5)\\ =\gcd(-5x-3, x+5)\\ =\gcd(-5x-3+5(x+5), x+5)\\ =\gcd(22, x+5)\\ =\gcd(1, x+5)=1 $$

Arthur
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  • Sorry, how come 5x is positive, not negative? My final polynomial was 22, not -22. – CoopDaddio Apr 22 '19 at 06:50
  • I realised I am getting 22 instead of -28... Where is my mistake? – CoopDaddio Apr 22 '19 at 06:54
  • @CoopDaddio Yes, sorry, that was indeed my mistake. I had both $x+5$ and $x-5$ in my head for some reason and got them mixed up. – Arthur Apr 22 '19 at 06:56
  • So why is 22 equivalent to 1? And does this mean all polynomials that are just integers are all equal to 1 in this case? – CoopDaddio Apr 22 '19 at 06:58
  • Sorry just want to confirm before you answer: so if the divisior of 2 polynomials that are non-zero and have a single variable (x in this case) is ever just an integer value (either negative or positive), and the remainder is 0, the GCD is always 1. Is this correct? – CoopDaddio Apr 22 '19 at 07:27
  • @CoopDaddio If you reach a non-zero, constant polynomial doing the Euclidean algorithm on polynomials with real coefficients, then yes, the GCD is $1$. You can see the same effect at play when finding the GCD of two integers and the result is $-1$. That doesn't mean that the GCD is $-1$, but rather that $1$ and $-1$ are equivalent when it comes to divisibility (what one divides, the other one divides too), and so you swap one for the other. With integers, we like our GCD to be positive, with real polynomials we like our GCD to be monic. – Arthur Apr 22 '19 at 10:40
  • That being said, if you are working with polynomials with integer coefficients, rather than real or rational, then the Euclidean algorithm does fail the moment you reach $\gcd(x+5, 22)$, precisely because you can't get any further, but you're not finished. If you ever get to work with ring theory, attempting to fix this discrepancy was, if i recall correctly, the idea behind what we call ideals, and the reason they have that name. – Arthur Apr 22 '19 at 10:43