Recent content by AcousticBruce

This is exactly what a friends said. The size gap. Ill think on this the rest of the day and see if I can find why that matters. Thanks.

Prove that the product of four consecutive natural numbers cannot be the square of an integer. So let n be a natural number. So f(n) = n(n+1)(n+2)(n+3) n --- 1 --- 2 --- 3 --- 4 --- 5 --- 10 f(n)-24--120-360-840-1080-17160 The conjecture I want to prove is F(n) + 1 is always a square...

That is because I did it wrong. :/ ## {\frac{{ -(1) \pm \sqrt {(1)^2 - 4(1)(1)} }}{{2(1)}}} ## I believe these are the answers. ## -\frac{1}{2} + \frac{\sqrt{3}}{2}\cdot i## ## -\frac{1}{2} - \frac{\sqrt{3}}{2}\cdot i##Edit:: Is there another way to figure this out without the quadratic formula?

Man.. I need another hint. I used the quadratic formula with ## y^2 + y + 1 ## and got ## -1-i ## and ## -1+i ##. Am I at least going in the right direction?

Well, I know that ## i^2 = -1 ## and ## i^4 = (-1)(-1) = 1 ##. I am just unaware how to make the polynomial equal zero. ##a^4 + a^2 b^2 + b^4## factor out ##b^4## ## b^4(\frac{a}{b})^4+(\frac{a}{b})^2+1) ## convert ##\frac{a}{b}## to ## x ## ## x^4+x^2+1 ## Hopefully you do not have to...

## a^2 - 6ab + 8b^2 ## this factors pretty easily to ## (a-2b)(a-4b) ## But that is not what I was wanting to learn... so Ill start again with the roots method. ## a^2 - 6ab + 8b^2 ##3 I will factor out ## b^2 ## ## b^2((\frac{a}{b})^2-6(\frac{a}{b})+8) ## make it simple with ## x^2-6x+8...

I really appreciate all of the effort put into explaining this too me! Would you are anyone else here mind giving me another problem that works out similar to this one so I can practice? a^3 - 6 a^2 b + 11ab^2 -6b^3

This is exactly what I do not understand. Wher does the -1 -2 and -3 come from. Thank you

I am completely confused by the -1 -2 and the -3. Can you help me understand this Micromass?

I am still studying this and I will sleep on it and delve into it excitedly tomorrow. I am loving this. Thank you so much. I agree. This is the much easier route. Instead of going for a cube problem first, it could have been easily the difference of squares, this would have left me 2 separate...

It took me a while to see it, but now I see it. Geez, this is crazy. I might not have ever figured this out. So when you say "forced" you mean this is not normal? How is subtracting ##a^2b^2## not clever? This reminds me of completing the squares, but it is so different then the problems I am...

I am sorry. I have no idea what you mean. How is ## (a^4 + a^2y^2 + b^4) ## a difference of squares?

Hello folks. I pulled out my algebra and trigonometry book that I kept from college (that I never ended up going). I am brushing up with algebra right now and this is something that stumped me. If you do not mind, I would love to learn how I can think differently in order to complete this...

Wonderful! Thank you both for the time and effort not only does this work, but I understand it completely. I also added a way to change the percentage on the calculator. Wonderful!

The problem is we have 2 unknowns: Transaction fee and Total Price. Fee can be derived from Total Price. Total price can be derived from Fee + Cost + Profit We only have Cost and Profit Basically, I could write a brute force method in Java, but I just think there is some math I do not...