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For the second example, if you use the same a=mb method, (m^2)(b^2) - mb^2 + b^2 (b^2)(m^2 -m +1) Ie. m=1 is a solution for any b And we can ask at what values multiplying b^2 by (m^2 -m +1) produces a square. Well, all i could see is where b^2 = m^2 - m +1 b^2 -1 = m^2 -m And without going further my gut says there are no more answers here (thinking about it as factors, I can't see another positive integer solution, but I'm not math-smart; i don't know why I'm watching this). Anyway, Soln: a = b or b=0. But the b=0 answer seems trivial. Especially as it works for the first question too as people have pointed out. Having a variable as zero just removes it and makes the formulation of the question redundant. Which is why i thought you avoided it. but then the second question it's what you gave.
So there are so far two types of solutions, [b=0 and a = any integer other than zero] and [(a,b) = (2,1) and (4,2),... and (2n,n)]. But are there other types of solutions, or can it be proven that there are not?
FAIL!! M=2 is not the only solution. What about when b=0, or m 2 or any form other than (mb,b)? Why did you start with (mb,b), was it just a guess? (a,0) and (a,-a) are also solutions. I can't prove there aren't more. That's why I watch videos like this one. This gets zero on a quiz, part marks at best.
The point was to find how many solutions there are. He proved that there are infinitely many. The fact there's other forms won't change that cardinality
I'll be traveling for Thanksgiving soon, but I've tried to get ahead so daily videos will continue. Some quicker and easier discussions but I hope you'll find them fun nonetheless!
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Let a be an arbitrary nonzero integer and b equal to 0. Then (a³+b³)/(a-b) = a³/a = a².
6:17 b=0 a≠0 also works for the first question, it's just that it doesn't fit (mb, b) but (a³+0)/(a-0) =a²
I wove this guy's dry humor "... and let a be anything whatsoever" is so heat
dude you pump out videos faster than i can possibly watch them. in terms of math channels you are number 1 in output
I'm always grinding, got a fun paradox video lined up for tomorrow!
For the second example, if you use the same a=mb method,
(m^2)(b^2) - mb^2 + b^2
(b^2)(m^2 -m +1)
Ie. m=1 is a solution for any b
And we can ask at what values multiplying b^2 by (m^2 -m +1) produces a square. Well, all i could see is where
b^2 = m^2 - m +1
b^2 -1 = m^2 -m
And without going further my gut says there are no more answers here (thinking about it as factors, I can't see another positive integer solution, but I'm not math-smart; i don't know why I'm watching this). Anyway,
Soln: a = b or b=0.
But the b=0 answer seems trivial. Especially as it works for the first question too as people have pointed out. Having a variable as zero just removes it and makes the formulation of the question redundant. Which is why i thought you avoided it. but then the second question it's what you gave.
i actually did the exact same thing as you and got the solution where m = 1. funny because our working out is basically exactly the same.
I love the nintendo music in the background
Classic tunes!
@@WrathofMathwhat's the song at the end right before the credits?
So there are so far two types of solutions, [b=0 and a = any integer other than zero] and [(a,b) = (2,1) and (4,2),... and (2n,n)].
But are there other types of solutions, or can it be proven that there are not?
Shakespeare would have liked this problem... 2b or not 2b. 😜🤣
huh, but what about other pairs? Do any other pairs work? I wonder how hard this problem is
√91
√(1001/9
U
I am so early
Great Video btw
Thank you!
FAIL!! M=2 is not the only solution. What about when b=0, or m 2 or any form other than (mb,b)? Why did you start with (mb,b), was it just a guess? (a,0) and (a,-a) are also solutions. I can't prove there aren't more. That's why I watch videos like this one. This gets zero on a quiz, part marks at best.
Oh so are there more than infinity answers?
The point was to find how many solutions there are. He proved that there are infinitely many. The fact there's other forms won't change that cardinality
yes, steve, you failed in comprehensivly reading the problem.