Question

If (z + 1/z)2 = 98, compute (z2+ 1/z2) ?
Can you please share the exact property that needs to be applied here? I am getting the wrong answer. It'd be helpful if you left a Khan Academy link for the topic as well.

Answer 1

Answer:

Multiply both sides by x2y2z2.

Then you get

y2z2+x2z2=x2y2

Now use that each of x2,y2,z2 divide two of the terms hence the third.

Added Here is the rest of the solution. Let a=gcd(x,y),b=gcd(x,z),c=gcd(y,z).

Then, gcd(a,b)=1 and hence ab|x. We claim ab=x.

Indeed write x=abd. Assume by contradiction that d≠1 and let p|d, p prime.

As x|yz we have abd|yz⇒d|yazb.

Then p divides either ya or zb.

But then, in the first case pa|x,y while in the second pb|x,z contradicting the gcd.

Therefore x=ab. The same way you can prove that y=ac,z=bc.

Replacing in the above equation you get

a2b2c4+a2b4c2=a4b2c2

or

c2+b2=a2

this shows that (c,b,a) is a primitive Pytagoreal triple and

x=aby=acz=bc