Let a, b be positive integers and set d = Gcd(a, b).

(a) Prove d|Gcd((a^2) + (b^2), a + b).

(b) If a and b are odd, then prove 2d|Gcd((a^2) + (b^2), a + b).

I tried to solve it but I couldn't

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- Apr 29th 2009, 09:11 AMsmithhallAlgorithim
Let a, b be positive integers and set d = Gcd(a, b).

(a) Prove d|Gcd((a^2) + (b^2), a + b).

(b) If a and b are odd, then prove 2d|Gcd((a^2) + (b^2), a + b).

I tried to solve it but I couldn't - Apr 29th 2009, 02:21 PMZeroDivisor
Since d divides both a and b, there are integers k and l s.t.

a = kd and b = ld.

Thus a^2 + b^2 = (k^2+l^2)d^2

and a+b = (k+l) d

Thus, the gcd of a^2+b^2 and a+b is at least d, which is a)

Now, for b), we assume that a and b are both odd.

Then k and l are odd, too.

Thus k^2+l^2 is even (odd + odd is odd and odd + odd is even)

and k+l is even.

Thus 2d divides the gcd of a^2+b^2 and a+b in this case.

Best,

ZD