Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One thing which i got is $m^4 + 4^n$ is congruent to $1 \pmod 8$ when both $n,m$ are odd...
Is it an iff condition?

share|cite|improve this question
Please check whether my edit did not unintentionally change the meaning of your question. – P.. Jan 15 '13 at 12:47

$1^4+4^1=5$ is prime, but n is not even.

$m=1$ and $n=6$, $1+4^6=17\times 241,$ so is not prime.

share|cite|improve this answer
sorry m not equal to 1 – Mambo Jan 15 '13 at 12:34
@user49236 : what about $n=0$? – Henry Jan 15 '13 at 12:45
@user49236 I just edited the answer. – ՃՃՃ Jan 15 '13 at 12:54
$3^4+4^9=5^2\cdot17\cdot617$ – P.. Jan 15 '13 at 12:55
Sorry, I made a mistake. – ՃՃՃ Jan 15 '13 at 12:55

It follows easily from $$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2 xy+2 y^2) \ . $$

share|cite|improve this answer
If $n=2k+1$ take $x=m$ and $y=2^k$ – PAD Jan 15 '13 at 13:09

There are primes when $m=1$ or when $n=0$ (indeed all but one of them are the same). You are probably excluding them.

For $n=1$ you have $m^4+4^1 = ((m-1)^2 +1) \times ((m+1)^2 +1)$ which is not prime unless $m=1$.

For $n \gt 1$ you have $m^4+4^n \equiv 0 \text{ or } 1 \mod 8$, which leads to your question

share|cite|improve this answer

If you make some arithmetic modulo $\,4\,$:

$$m^4+4^n=m^4=0,1\pmod 4\Longrightarrow \,\,m\,\,\text{must be odd (why?)}$$

Also, if

$$n=2k+1\Longrightarrow 4^n=4\cdot (4^k)^2\,$$

and Pantelis's answer kicks in.

share|cite|improve this answer

Your Answer


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.