Prime Powers

Number Theory Level 4

$\begin{cases} n=2^a + 1 \\ n = a^2 + 2^b \\ n = c^2 -2^{b+1} \end{cases}$

If $a,b$ and $c$ are prime numbers that satisfy the system of equations above, find $n$ .

The answer is 33.

1 solution

Chew-Seong Cheong
Mar 15, 2016

From $n = 2^a + 1 = a^2+2^b$ , we have:

$\begin{aligned} 2^a + 1 & = a^2 + 2^b \\ \Rightarrow 2^b & = 2^a + 1 - a^2 \quad \quad \small \color{#3D99F6}{\text{As the LHS is even, } a \text{ must be odd.}} \end{aligned}$

For $a=3$ , we get $LHS = 2^3 + 1 - 3^2 = 0$ not acceptable. For $a=5$ , we have:

$\begin{aligned} 2^b & = 2^5 + 1 - 5^2 \\ & = 32 + 1 - 25 \\ & = 8 \\ \Rightarrow b & = 3 \end{aligned}$

Let us check if $a=5$ and $b=3$ is $c$ and prime.

$\begin{aligned} n & = 2^5 + 1 = 33 \\ \Rightarrow 33 & = c^2 - 2^{3+1} \\ \Rightarrow c^2 & = 33 + 2^4 = 49 \\ c & = 7 \text{ (a prime)} \end{aligned}$

$\Rightarrow n = \boxed{33}$

How to prove that the rest of the prime numbers do not satisfy the given equation (not because the question is asked for single value of 'n')

Bala vidyadharan - 5 years, 3 months ago

Of course, but I have no proof for it.

Chew-Seong Cheong - 5 years, 3 months ago

Prime Powers

The answer is 33.

1 solution

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