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# 199

Happy Saint Math Trick's Day! - March 17, 2010
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Am I missing something in here? (x^2+14)/12=3 solutions are x = -sqrt(22) and x = sqrt(22), far from being primes.

Remainder... I should learn to read.

Makes sense now. Integer solutions for x^2+14 mod 12 = 3 are x = 6n+1 and x = 6n+5.

6n+2 is divisible by 2, 6n+3 by 3, 6n+4 by 4, 6n by 6 (and 2 and 3).

Step 1: Drink 10 beers.
Step 2: Prove the remainder is always 3.
Step 3: Drink more beer.

You can't count 'beer' (which means, you can't say 10 beers)
because 'beer' is Uncountable Noun :D

I think you should say 'A glass of beer'

That's only true if you've either drunk too many beers.

It could also refer to different kinds of beer.

I once had uncountably many beers.

x*x+14=y*12+3
where y is an integer
doesn't work for x = 7919, though ;)

It does work.

oh shoot, it does, forgot to add 14 :(
at least now I see why it works ;)

I can't make it work...
5^2= 25. 25+14= 39. 39 divided by 12... isn't even... :-(

39/12=3 R3
aka
12*3=36+3=39

The remainder is what is left. It doesn't have to divide evenly.

wtf 3? It always give me .25

.25 = 1/4 = 3/12

Verify it by the algorithm you used in elementary school.

If x is a prime greater than 3, then x is congruent to 1, 5, 7, or 11 modulo 12. In all of this cases, x^2 = 1 mod 12. Adding 14 to both sides of the congruence yields x^12 + 14 = 15 = 3 modulo 12.

Good stuff.

Man...I'm an idiot...

1^2 -> 1/12 = 0.083~
5^2 -> 25/12 = 1.083~
7^2 -> 49/12 = 4.083~
11^2 -> 121/12 = 10.083~

Add 14 before dividing by 12!
Then the decimal part will be .25 which is 3/12 so the trick works

Nicely put. Perhaps moreso, given that that's exactly the logic I used to derive it ;-)

If you don't understand the meaning of "remainder", or if you can't read English, please just save your breath and don't embarrass yourself criticizing the author. :(

Agreed. I don't understand how someone could be a fan of spikedmath without understanding remainders. I wonder how many other strips go completely over their head.

701 * 701 = 491401
491401 + 14 = 491415
491415 / 12 = 40951.25 -> it works!

Use the .25 as your verification as 3/12 = 1/4 = .25

prime>3 is always 6n+1 or 6n-1
square it: 36n^2 +/- 12n + 1
add 14: 36n^2 +/- 12n + 15
mod 12: =3

It works for every odd number not dividable by 3.
And for odds dividable by 3 the outcome equals 11.

And here I was thinking this could be used to proving the Reinmann Hypothesis. Damn it!

I guess you meant the Riemann Hypothesis...

Reading the dialog is often funnier than the comics...

Lol yeah :D

Awesome, LOL

Do only primes work for this?

Any number that doesn't have 2 or 3 as factors works. so, 25,35,etc. (not prime) also work.

thsts jst a straighforward application of fermat's little theorem

You can push this even further.
Pick a prime greater than 5.
calculate it's 4th power.
Your remainder will be: *drumroll* 3.
Works precisely the same way the easier formula from the comic works.

For anyone who wants to know how this works and couldn't or wouldn't understand any of the previous posts, here's what I worked out:
p^2+14 mod 12 = 3 is basically a formal restatement of what's done in the comic.
Now this can be simplified a bit:
p^2+14 mod 12 = 3
=> p^2 mod 12 +14 mod 12 = p^2 mod 12 + 2 = 3 (14 mod 12 = 2)
=> p^2 mod 12 = 1
Some people are uncomfortable with the "mod" notation, so lets rewrite that:
p^2 = 12n + 1 (for some integer n)
=> p^2 - 1 = 12n
=> (p-1)(p+1) = 12n (left side factorized)

Now lets look at what we have got here. p is a prime greater than 3, so both (p-1) and (p+1) are even numbers (divisible by 2). Since they are 2 consecutive even numbers, one of them is even divisible by 4.
Let's consider the 3 consecutive numbers (p-1),p,(p+1). In 3 consecutive numbers, precisely 1 is divisible by 3. Since p>3 and prime, p will not be divisible by 3; thus either (p-1) or (p+1) is divisible by 3.
Hence the product (p-1)(p+1) is guaranteed to have the factors 2, 3, and 4 so 2*3*4=24 divides (p-1)(p+1) without remainder. A number that is divisible by 24 can be written in the form 12*n (for some integer n).
It follows that
p^2-1=12n
and
p^2 + 14 mod 12 = 3

We didn't really use the fact that p is prime, only that p is an odd number that doesn't have 3 as a factor. So any number that fulfills these requirements will work, not only primes (has already been remarked upon in the previous comments).

Btw., if you employ the same method to the p^4+722 mod 240 = 3 problem you will get pretty much the same basic results, only more factors have to be deduced.

according to Chinese remainder theorem, remainder mod 12 is uniquely determined by the remainders mod 3 and 4.
3 mod 12 is obviously 0 mod 3 and 3 mod 4.
mod 3: p!=0 which means p^2=1 by Fermat's little theorem, so p^2+14=0.
mod 4: gcd(p,4)=1 which means p^2=1 by Euler's theorem, so p^2+14=3.
QED

P.S. using this form of proof the argument can be generalized to bigger primes.

It's not that difficult to prove. No beer was consumed in the making of this proof.

Adding 14 is another way to say adding 2 in mod 12. But if you subtract 3 more, then the number should have no remainder. (same logic)

Call the prime p. We currently have p squared - 1, or (p+1)(p-1)
Any prime number greater than 2 can be expressed as 2k+1, where k is a natural number.
Thus, we have (2k)(2k+2), which can be manipulated into 4(k)(k+1), which means that (p+1)(p-1) is divisible by 4.
(p+1)(p-1) is divisible by 3 as well. Put the number p in between. For any 3 consecutive integers, at least one of them is divisible by 3. Since p is a prime number and can't be divisible by 3 unless it is 3, either p+1 or p-1 is divisible by 3. That's all we need. If it's divisible by 3 and 4, it is divisible by 12.

Done!

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