<p>Oh okay. This is really cool.</p>
<p>Is an empty set considered a minimal selfish subset?</p>
<p>^ The empty set is not selfish. It contains 0 elements but 0 is not an element of the empty set. </p>
<p>Set theory is headache inducing…</p>
<p>^ To make it more headache inducing: in some variants of set theory (at least in what I was taught), the number 0 is defined as the empty set! As in literally, the symbol 0 means {}. Then they define 1 = {0} = {{}}, and 2 = {0,1} = {{}, {{}}} and so on. My teacher said it was ‘naive’ set theory! I can’t even imagine what, uh, un-naive set theory looks like. </p>
<p>These are good questions! Still, this seems like it’d be more like a problem marathon on AoPS than a thread on CC; knowing how to do this kind of stuff isn’t necessary in high school if you’re not doing the AMC. Interesting and enlightening, though!</p>
<p>oh… {1} is a minimal selfish subset, right?
This clears some things…</p>
<p>
</p>
<p>STOP! Haha… This is really… just too much 
I don’t think this thread has anything to do with SAT prep, but I like the thinking… (although I’m not exactly good at it).</p>
<p>@Jeffrey: yes, {1} is minimal selfish. So is {2,n} for n>2 as another list of examples.</p>
<p>I didn’t expect people to get solutions to these so quickly. Great job so far. Let’s see how you do with this one:</p>
<p>Let n be a positive integer such that n>2. Show that n^4-6n^2+1 is not prime.</p>
<p>This is a good one; I’ve seen it before so I’ll wait for others to answer. (they ought to ask these kinds of questions in math class! never seen anything like it in school though.)</p>
<p>After this one is solved, here’s a similar one for more practice: prove that n^4 + 64 is not prime either.</p>
<p>Can you give me a hint?
I know that n^4-6n^2+1 always yields a multiple of 2 when n = odd integer…</p>
<p>(2k)^4 - 6(2k)^2 + 1
16k^4 - 24k^2 + 1 = (4k^2-4k-1)(4k^2+4k-1)</p>
<p>Can I conclude that since n^4-6n^2+1 is factored by (4k^2-4k-1) and (4k^2+4k-1), it is not prime?</p>
<p>I’m confused…!</p>
<p>Okay. I cheated… in the last post… with a calculator… 
But it helped me figure out what to do…</p>
<p>n^4-6n^2+1
n^4-2n^2+1-4n^2
(n²-1)²-(2n)²; difference of two squares</p>
<p>(n²-1-2n)(n²-1+2n) = factors of n^4-6n^2+1</p>
<p>The idea of completing a square and then using difference of squares on the result is sometimes called the Sophie Germain Identity:</p>
<p>[Sophie</a> Germain Identity - AoPSWiki](<a href=“http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie”>Art of Problem Solving)</p>
<p>Of course, it’s much more valuable to know how it works rather than the formula they give. DrSteve doesn’t seem to be here at the moment, so I guess I’ll add another one.</p>
<p>If you choose two random real numbers between 0 and 1, what’s the expected value of the smaller one? (extension: what if you choose n numbers?)</p>
<p>I learned about this method of adding and subtracting in Pre-Calculus. I didn’t really know the name of this, but I like the name! </p>
<p>I don’t think I understand the question… Can you elaborate? What do you mean by “expected value”?</p>
<p>
</p>
<p>These are obviously much harder than SAT problems, but I think they’re fun to try, and attempting to solve difficult problems leads to an increase in mathematical maturity which indirectly increases SAT scores. Struggling with these problems for a while will make those Level 5 SAT problems seem like a breeze. </p>
<p>If people think that this thread is inappropriate for this forum I’m happy to drop it, and maybe come up with something easier, but I personally don’t think it’s a bad thing for students here to be working on difficult math problems.</p>
<p>
</p>
<p>You cannot conclude this yet. If you can show that neither of those factors are 1 or -1, then you can conlcude that n^4-6n^2+1 is not prime.</p>
<p>In fact, if you plug in 2 for n, you will see that you do get a prime. This is why I put the condition n>2.</p>
<p>
</p>
<p>Again, you still have to show that neither of those factors are 1 or -1.</p>
<p>
</p>
<p>Yes… Under the condition n>2, the factor will never be 1 or -1… right?</p>
<p>
</p>
<p>No…no!!! I really enjoy the problems you give.</p>
<p>
</p>
<p>That’s correct. Can you provide the details? It would be nice to have a complete solution for others that are reading this thread.</p>
<p>(n²-1-2n)(n²-1+2n);</p>
<p>I don’t really know how to explain this… but for n²-1-2n, n>2 is the point in which the result is greater than 1…</p>
<p>How can I prove this?</p>
<p>I don’t want this thread to die!</p>
<p>n^4-6n^2+1
The result must be greater than 1: n>1
(n²-1-2n)(n²-1+2n) :</p>
<p>n²-2n-1 ≠ -1;
n²-2n ≠ 0
n(n-2) ≠ 0
n ≠ 0 and 2
n>2</p>
<p>Is this correct?</p>
<p>Here, you can prove it’s bigger by completing the square:</p>
<p>n^2 - 2n - 1 = (n-1)^2 - 2 > 1, so n-1 > sqrt(3) so n > sqrt(3) + 1. That’s about 2.7, so letting n > 2 for integer n works. You could also prove it by taking the derivative, but that’s no fun. More or less the same thing works for the other factor.</p>
<p>Can you offer the derivative solution? I would love to see it… :)</p>
<p>
</p>
<p>What you showed here is that if n²-2n-1 = -1, then n=0, or n=2. </p>
<p>If you want to finish the problem this way you need to solve 3 more equations by setting this factor equal to 1, and the other factor equal to each of 1 and -1.</p>
<p>To summarize, we need to show that if n is an integer greater than 2, then each of those factors is not equal to 1 or -1. This is easy. We set each factor equal to 1, and to -1, and solve the 4 resulting equations. The solutions are 0,2, and the rest are not integers (check this using the quadratic formula, for example).</p>
<p>
</p>
<p>I’m not sure what you did here, but I don’t think it works. For example, if n=0, then n^2 - 2n - 1=-1 which is not greater than 1.</p>
<p>
</p>
<p>I’m not sure if this is what conqueror had it mind, but you can do the following:</p>
<p>The derivative of f(n)=n^2-2n-1 is f`(n)=2n-2, which is 0 when n=1. To the left of 1, the derivative is negative, and to the right of 1 the derivative is positive. So f is strictly decreasing until 1, and strictly increasing after 1. f(1)=-2, f(0)=f(2)=1, and by what the derivative showed about f, f(n)>1 for all integers n greater than 2, or less than 0.</p>
<p>The argument for n^2+2n-1 should be similar.</p>