Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that :

If $a^n+n^{a}$ is prime number and $a=3k-1$ then $n\equiv 0\pmod 3$

where $a>1,n>1 ; a,n,k \in \mathbb{Z^+}$

I have checked statement for many pairs $(a,n)$ and it seems to be true.

Small Maple code that prints $(a,n)$ pairs :

enter image description here

Any idea how to prove this statement ?

share|improve this question

2 Answers 2

Since it is prime, we know that $a^n+n^a \equiv 1 \text{ or } 5 \mod 6$, and because $a \equiv 2 \mod 3$ we know that $a \equiv 2 \text{ or } 5 \mod 6.$

The case $a \equiv 2 \mod 6:$
Note that $n^2 \equiv n^8 \equiv n^{14} \dots \mod 6.$
When $n \equiv 0, 1, 2, 3, 4, 5 \mod 6, n^2 \equiv 0, 1, 4, 3, 4, 1 \mod 6$ and $2^n \equiv 4, 2, 4, 2, 4, 2 \mod 6.$
Adding these gives $a^n+n^a \equiv 4, 3, 2, 5, 2, 3 \mod 6$. 1 does not appear at all and 5 appears only when $n \equiv 3 \mod 6,$ meaning that $n \equiv 0 \mod 3,$ as required.

The case $a \equiv 5 \mod 6:$
Note that $n^5 \equiv n^{11} \equiv n^{17} \dots \mod 6.$
When $n \equiv 0, 1, 2, 3, 4, 5 \mod 6, n^5 \equiv 0, 1, 2, 3, 4, 5 \mod 6$ and $5^n \equiv 1, 5, 1, 5, 1, 5 \mod 6.$
Adding these gives $a^n+n^a \equiv 1, 0, 3, 2, 5, 4 \mod 6$. 1 appears only when $n \equiv 0\mod 6$, implying that $n \equiv 0 \mod 3,$ as required.
However, 5 appears as well, when $n \equiv 4 \mod 6,$ suggesting the possibility of the result being false for these values of $a$ and $n$.

Edit
I have found that $a=215$ and $n=76$ is a counterexample. $a^n+n^a$ is prime, $a \equiv 2 \mod 3$ but $n\not\equiv 0 \mod 3$.

share|improve this answer
    
if $n\equiv 0\pmod 3 $ then $n \not\equiv 4\pmod 6$ –  pedja Nov 17 '11 at 15:18
    
@pedja I have shown that $a^n+n^a$ might be prime without the need for $n \equiv 0 \mod 3$. $n \equiv 0 \mod 3$ is your conclusion, not an assumption. –  Peter Phipps Nov 17 '11 at 15:23
    
could you give me an example of such prime number please ? –  pedja Nov 17 '11 at 15:30
    
@pedja that would be the counterexample to disprove your conjecture. Your Maple code suggests you have already checked the easy numbers. So no, I can't provide an example at this time. –  Peter Phipps Nov 17 '11 at 15:36
    
if you find counterexample to my conjecture please let me know.. –  pedja Nov 17 '11 at 15:40

Assume $n$ ist not $\equiv 0 \mod 3$, which means $n=3l+1$ or $n=3l+2$. Now see if $n^a + a^n$ is divisibile by 3. If yes, you get a contradiction to $n^a + a^n$ being a prim.

So if $n \equiv 1$ you get (remember $a \equiv -1$) $$a^n + n^a \equiv (-1)^{3l+1} + 1 \equiv 0 \mod 3$$ and if $n \equiv 2$ you get $$a^n + n^a \equiv (-1)^{3l+2} + (-1)^{3k-1} \equiv 1-1 = 0 \mod 3$$

share|improve this answer
    
counterexamples : $8^2+2^8=320 , 8^4+4^8=69632 $ –  pedja Nov 16 '11 at 9:34
    
Dont you assume that $a^n + n^a$ is prime ? –  MaoK Nov 16 '11 at 9:37
    
try $2^{15}+15^2$ –  pedja Nov 16 '11 at 9:52
    
Yap, thats a prim. But thats no counter example, since $n=15 \equiv 0 \mod 3$ –  MaoK Nov 16 '11 at 9:59
    
please calculate : $2^{14}+14^2$ and $2^{16}+16^2$ –  pedja Nov 16 '11 at 10:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.