Am I missing something in here? (x^2+14)/12=3 solutions are x = -sqrt(22) and x = sqrt(22), far from being primes.

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

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... :-(

wtf 3? It always give me .25

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.

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. :(

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

So prime is somewhat misleading...
It works for every odd number not dividable by 3.
And for odds dividable by 3 the outcome equals 11.

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

If all else fails, return to Euler.

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.
Add 722. Divide by 240.
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.