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First, we can prove that in order to make the nonzero difference between two perfect squares as small as possible, their square roots must differ by 1. We can prove this as follows:

Let the two squares be \(a^2\) and \(b^2\). Without loss of generality, let us assume \(b \geq a, \) so we can write \(b \) as \(a+k.\) Now, to compute the difference, we get \(b^2-a^2=(a+k)^2-a^2=a^2+2ak+k^2-a^2=2ak+k^2.\) Obviously, in order to minimize the nonzero difference, \(k \neq 0, \) and \(k \) must be as small as possible.

Because we want \(2ak+k^2 \) to be 1, we can factor this as \(k(2a+k)\) to see that either \(k=2a+k=-1,\) or \(k=2a+k=1.\) 

If  \(k=2a+k=-1,\)  \(2a-1=-1,\) so \(a=0,\) which is a contradiction.

If \(k=2a+k=1,\) \(2a+1=1,\) so \(a=0,\) which is a contradiction.

Thus, two nonzero squares \(a^2\) and \(b^2\) cannot differ by 1.

There seems to be missing text, can you fix this please?

If c and d are inversely proportional, then c*d always stays the same. c=9 when d=8, so c*d is always 72. This means that d=72/8=9 when c=8.

Thank you all so much for the helpful answers!!!!

Here is the solution given:

Let \(q(x) = (x^2 - 1) p(x) - x.\) Then \(q(n)\) has degree 7, and \(q(n) = 0\) for n=2, 3, 4, ... 7, so \(q(x) = (ax + b)(x - 2)(x - 3) \dotsm (x - 7)\)
for some constants \(a\) and \(b.\)

We know that \(q(1) = (1^2 - 1)p(1) - 1 = -1.\) Setting \(x=1 \) in the equation above, we get \(q(1) = 720(a + b),\)
so \(a+b=-\frac{1}{720}.\)

We also know that \(q(-1) = ((-1)^2 - 1)p(-1) + 1 = 1.\) Setting \(x=-1\) in the equation above, we get \(q(-1) = 20160(-a + b),\)
so \(-a + b = \frac{1}{20160}.\) Solving for \(a\) and \(b,\) we find \(a = -\frac{29}{40320}\) and \(b = -\frac{3}{4480}.\) Hence, \(\begin{align*} q(x) &= \left( -\frac{29}{40320} x - \frac{3}{4480} \right) (x - 2)(x - 3) \dotsm (x - 7) \\ &= -\frac{(29x + 27)(x - 2)(x - 3) \dotsm (x - 7)}{40320}. \end{align*}\)
In particular, \(q(8) = -\frac{(29 \cdot 8 + 27)(6)(5) \dotsm (1)}{40320} = -\frac{37}{8}, \)
so \(p(8) = \frac{q(8) + 8}{8^2 - 1} = \boxed{\frac{3}{56}}.\)
 

If these numbers are correct, then what is p(8)?

First of all that's not an answer

Second of all, even the lyrics are wrong...

It's right!!!! Here is the solution provided on the website:

We can assume that the ellipse is tangent to the circle \((x - 1)^2 + y^2 = 1.\) From this equation, \(y^2 = 1 - (x - 1)^2.\) Substituting into the equation of the ellipse, we get \(\frac{x^2}{a^2} + \frac{1 - (x - 1)^2}{b^2} = 1.\) This simplifies to \((a^2 - b^2) x^2 - 2a^2 x + a^2 b^2 = 0.\)
By symmetry, the \(x\)-coordinates of both tangent points will be equal, so the discriminant of this quadratic will be 0: \((2a^2)^2 - 4(a^2 - b^2)(a^2 b^2) = 0.\)
This simplifies to \(a^4 b^2 = a^4 + a^2 b^4.\) We can divide both sides by \(a^2\) to get \(a^2 b^2 = a^2 + b^4.\) Then \(a^2 = \frac{b^4}{b^2 - 1}.\)
The area of the ellipse is \(\pi ab\) Minimizing this is equivalent to minimizing \(ab\), which in turn is equivalent to minimizing \(a^2 b^2 = \frac{b^6}{b^2 - 1}.\)
Let \(t=b^2\), so \(\frac{b^6}{b^2 - 1} = \frac{t^3}{t - 1}.\)
Then let \(u=t-1.\) Then \(t=u+1,\) so \(\frac{t^3}{t - 1} = \frac{(u + 1)^3}{u} = u^2 + 3u + 3 + \frac{1}{u}.\)
By AM-GM,

 \(\begin{align*} u^2 + 3u + \frac{1}{u} &= u^2 + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{u}{2} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} + \frac{1}{8u} \\ &\ge 15 \sqrt[15]{u^2 \cdot \frac{u^6}{2^6} \cdot \frac{1}{8^8 u^8}} = \frac{15}{4}. \end{align*}\)
Equality occurs when \(u=\frac{1}{2}\) For this value of \(u, t=\frac{3}{2}, b=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}, \) and \(a=\frac{3\sqrt2}{2}.\) 

Hence, \(k = ab = \boxed{\frac{3 \sqrt{3}}{2}}.\)

Whew! that took some time to format lol 

By the law of exponents, if \(3^a=2, \) and \(3^b=5,\) then \(3^{a+2b}=2 \cdot 5 \cdot 5 = 50.\) Now we know that \(50 \cdot 3 = 150, \) so \(3^{a+2b+1}=150,\) so in terms of \(a\) and \(b, \) \(\boxed{x=a+2b+1}\)

Hello Guest,

Here is my attempt:

There are 16 people who own cats, and 5 people who own cats and dogs, so there are 11 people who only own cats. Now there are 9 people who own dogs or both, so the answer would be \(11 + 9 = \boxed{20}\) people own a cat or a dog

First of all, thank you! 

But how do you know the foci must be at (-1,0) and (1,0)? Why does this minimize the area... 

Yes, I have. Here is the graph:

wouldnt k=1 work?

Thank you!!!! 

Use the Wolfram-Alpha Theorem to get x=-1/6

can you tell me how you did this?

Let us call a position where you can win if you play correctly a winning position, and any other position a losing one. We can see that having 1-7 left is a winning position, and 8 is a losing position, etc. Notice that whatever amount \(x\) that Morty takes on a given turn, Rick can always take \(8-x\) french fries, and thus 8 will be taken in total. This means that if Rick can turn the number of french fries into a multiple of 8, Morty will be in a losing position, and so Rick will win by just taking \(8-x\) french fries each turn, lowering the number by 8 each time until it reaches 0. In order to turn 60 into a multiple of 8, you need to take away 4, so the answer should be \(\boxed{b) \space 4}\)