Is $83^{27} +1 $ a prime number?

Question

I'm having problems with exercises on proving whether or not a given number is prime. Is $83^{27} + 1$ prime?

Answer 1

$83$ is odd, so is any power of $83$. Hence $83^{27}+1$ is even, but the only even prime number is $2$ and this number is not $2$.

More generally, if $a,k\in\mathbb N$ and $k$ is odd, then $$a^k+1\equiv (-1)^k+1\equiv 0\pmod{a+1}$$ So $a+1\mid a^k+1$. In this case this yields $84=2^2\cdot 3\cdot 7$ as divisor.

Answer 2

Well, it is an even number, so...

Answer 3

It is obviously not prime. $83$ is odd, therefore $83^{27}$ is odd, hence $83^{27}+1$ is even and not prime.

Answer 4

Note that $83\equiv -1\pmod{84}$. Thus $83^{27}+1\equiv 0\pmod{84}$.

It follows that our number is divisible by all the divisors of $84$.

It is also non-prime in other ways. For let $x=83^3$. Then our number is $x^9+1$, so is divisible by $x+1$. Similarly, we could let $y=83^9$, and conclude that our number is divisible by $y+1$.

Seriously non-prime!

Answer 5

$$ 83^{27} + 1 = \Big(83^9\Big)^3 + 1^3 = a^3+b^3 = (a+b)(a^2-ab+b^2) = \Big(83^9+1\Big)\Big((83^9)^2-83^9+1\Big). $$

So, no, it's not prime.

PS (added later): Some point out that it's obviously an even number, so it's not prime. But what I do above would work just as well if it were $84$ rather than $83$.

Answer 6

We have a chain of divisibilities, based on the fact that $(a-b)\mid(a^n-b^n)$, $$ 83^1-(-1)^1\mid83^3-(-1)^3\mid83^9-(-1)^9\mid83^{27}-(-1)^{27}=83^{27}+1 $$ Using this chain, we get, using $a^3-b^3=(a-b)(a^2+ab+b^2)$, $$ \begin{align} 83^{27}+1 &=\frac{83^{27}+1}{83^9+1}\times\frac{83^9+1}{83^3+1}\times\frac{83^3+1}{83^1+1}\times\left(83^1+1\right)\\ &=\left(83^{18}-83^9+1\right)\times\left(83^6-83^3+1\right)\times\left(83^2-83^1+1\right)\times\left(83^1+1\right)\\[9pt] &=34946659039493167203883141969862007\times326939801583\times6807\times84 \end{align} $$ Thus, $83^{27}+1$ is not prime.

Note: none of these factors are guaranteed to be prime, just factors.

Answer 7

The only prime numbers of the form $a^x+b^x$, occur when $x$ is a power of two. This does not guarantee a prime, but if $x$ is not a power of $2$, then the number has algebraic factors.

In practice, there is an algebraic divisor of $a^n-b^n$, for each $m$ that divides $n$. For the equation $a^n+b^n$, one would look for divisors of $2n$ that don't divide $n$. Inthe question we have $n=27$, so the divisors of 54 that don't divide 27. That is, 2, 6, 18 and 54. For powers of 2, there is only one number that divides $2n$ but not $n$.

Answer 8

Let's ask WolframAlpha!

PrimeQ[83^27 + 1]

is $6\,532\,937\,361\,590\,551\,025\,727\,805\,459\,013\,652\,074\,798\,022\,177\,030\,828$ a prime number?

$83^{27} + 1$ is not a prime number

$2^2 \times 3^4 \times 7 \times 109 \times 757 \times 2269 \times 9613 \times 49339 \times 2208799 \times 14685985270709080390792801 \space\text{(14 prime factors, 10 distinct)}$

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However, using basic knowledge that an odd times an odd is always an odd ($3 \times 3 = 9$), we see that $83$ (an odd number) raised to any power is an odd number. Then we add one to it and get an even number.

Being even (and obviously not equal to $2$), the definition of a prime tells us that the number is not prime because it is divisible by $2$ (my words):

prime (noun):

1. Any natural number, greater than $1$, that, when divided by any natural number, greater than $1$, other than itself or $1$ does not result in a natural number.
2. Any "natural number greater than $1$ that has no positive divisors other than $1$ and itself." (Wikipedia article "prime number")

Answer 9

well it is divisible by $84$ and in general $\forall a,m\in\mathbb {N}$ we have $(a+1)\mid (a^{2m+1}+1)$ So....