Algebra: Proving (a^2,b^2)=1 Using GCD Proof

maphec
Messages
3
Reaction score
0
(a,b)=d means d is the GCD of a and b

Question:

Let (a,b)=1

Prove: (a^2,b^2)=1



The hint that we were given is to prove this by contradiction ... but, I have no idea how to go about even starting this proof ... Any and all help would be greatly appreciated!
 
Physics news on Phys.org
Suppose that (a,b) = 1 and that (a2,b2) = k, where k > 1 ...
 
Proof:

Supposed n is prime and that n|ab
therefore, n|a or n|b, but not both

Case 1: n|a and n does not divide b
therefore n|a^2
since n does not divide b, n does not divide b^2

Case 2: n|b and n does not divide a
therefore n|b^2
since n does not divide a, n does not divide a^2

Thus, a^2 and b^2 have no common divisors
and therefore gcd(a^2,b^2)=1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top