Home > Uncategorized > Panaitopol primes – simple representation

Panaitopol primes – simple representation

A prime number {p} is called a Panaitopol prime if {\displaystyle p = \frac{x^4-y^4}{x^3+y^3}}. I stumbled upon these weird prime numbers while working on problem 291 from Project Euler.

While searching a bit about the problem I did a brute force check to find the first such primes and I found the following values:

\displaystyle 5, 13, 41, 61, 113, 181, 313, 421, 613, 761.

Since this was going nowhere I plugged these into the OEIS search engine and bam: it seems they are precisely the primes which can be written in the form {p=n^2+(n+1)^2}. If this latter formulation holds, then the problem can be solved immediately, since searching for primes of the form {n^2+(n+1)^2} is much easier than testing the first formula. Now how do we prove this? Suppose {\displaystyle p = \frac{x^4-y^4}{x^3+y^3}}, which is equivalent, after simplifying {x+y}, to
\displaystyle p = \frac{(x-y)(x^2+y^2)}{x^2-xy+y^2}\Rightarrow p(x^2-xy+y^2)=(x-y)(x^2+y^2).

Note that we necessarily have {x\neq y}. Now recall that {p} is prime, so if {p} divides {x-y} we must have {x^2+y^2|x^2-xy+y^2} which is false, since a positive number cannot divide a number smaller than itself (I leave you the task of seeing why {x^2-xy+y^2} cannot be zero). Therefore it remains that {p|x^2+y^2} and {x-y|x^2-xy+y^2}. Since {x-y|x^2-xy} and {x-y|y^2-xy} it follows that {x-y} divides {x^2,y^2} and {xy}. Therefore we can write
\displaystyle x^2=a(x-y),\ y^2=b(x-y),\ xy = c(x-y).

Multiply the first two relations and square the third to get {c^2=ab}. Furthermore, we have
\displaystyle (x-y)^2 = (a+b-2c)(x-y)

so {x-y = a+b-2c}. Now let’s see what remains of the initial {p}
\displaystyle p = \frac{(a+b-2c)(a+b)(x-y)}{(a-c+b)(x-y)}= \frac{(a+b-2c)(a+b)}{a+b-c}.

Now let’s suppose that {d=\gcd(a,b,c)} and denote {a_1=a/d,b_1=b/d,c_1=c/d}. Then we have
\displaystyle p = \frac{(a_1+b_1-2c_1)d(a_1+b_1)}{a_1+b_1-c_1}.

Now let’s note that if {q} is a common prime factor of {a_1+b_1-2c_1} and {a_1+b_1-c_1} then it follows immediately that {q} divides {x,y,a_1,b_1,c_1}, contradicting the choice of {d} above. Therefore we can assume that {\gcd(a_1+b_1+c_1,a_1+b_1-2c_1)=1}. This means that, for the above fraction to simplify we need to have {a_1+b_1-c_1|d(a_1+b_1)}. Furthermore, the ratio {\displaystyle \frac{d(a_1+b_1)}{a_1+b_1-c_1}} is strictly greater than {1}. Moreover, {\gcd(a_1+b_1,a_1+b_1-c_1)=1} so the previous fraction is in fact just {a_1+b_1}. Since {p} is prime, this forces the relation {a_1+b_1-2c_1=1}, which, since {c_1=\sqrt{a_1,b_1}}, gives us
\displaystyle \sqrt{a_1}-\sqrt{b_1}=1.

Therefore {a_1=(n+1)^2} and {b_1=n^2} for some positive integer {n}. In conclusion
\displaystyle p = a_1+b_1=n^2+(n+1)^2.

Now if we have {a,b,c} we can retrace the construction to find a the corresponding {x} and {y}. Note that in addition to {x-y = a+b-2c = d((n+1)^2+n^2-2n(n+1)) = d} we also have {x+y = a-b = d(2n+1)}. Choosing {d=n^2+n+1} we get exactly
\displaystyle \frac{x^4-y^4}{x^3+y^3}=n^2+(n+1)^2.

Categories: Uncategorized
  1. March 11, 2017 at 11:20 am

    nice analysis

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