Two To The Fiftieth Plus One

We know,

$a^{n+1}+b^{n+1}=(a+b)(a^{n}-a^{n-1}b+...+b^{n})$

Where $n$ is an even number.

And,

$2^{50}+1=(2^{10})^{5}+1^{5}$

Clearly it is not a prime.

2^{ 1 }\quad \equiv \quad 2\quad (mod\quad 10)\\ 2^{ 2 }\quad \equiv \quad 4\quad (mod\quad 10)\\ 2^{ 3 }\quad \equiv \quad 8\quad (mod\quad 10)\\ 2^{ 4 }\quad \equiv \quad 6\quad (mod\quad 10)\\ 2^{ 5 }\quad \equiv \quad 2\quad (mod\quad 10)\\ \vdots \\ then\quad 2^{ 50 }\quad \equiv \quad 4(mod\quad 10)\quad and\quad 2^{ 50 }+1\quad \equiv \quad 5(mod\quad 10)\quad therefore\quad was\quad a\quad multiple\quad of\quad 5

$(2^{50}+2(2^{25})+1)-2(2^{25}) = (2^{25}+1)^2-(2^{13})^2 = (2^{25}+2^{13}+1)(2^{25}-2^{13}+1)$ Hence proved that $2^{50}+1$ is not a prime.

It ends with a 5. The pattern for (2^n) ends in 2, 4, 8, 6... For n=1,2,3,4 and n=50 is only a multiple of n=2, therefore it ends in 4, and 4+1=5

If it were prime, the proof would be well beyond the scope of the question. Since the question is multiple choice, it must be composite.

Lazy programmer's solution:

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target = 2 ** 50 + 1
stop = int(target ** .5)

for i in xrange(2, stop):
    if target % i == 0:
        print('{} is divisible by {}'.format(target, i))
        break

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1125899906842625 is divisible by 5

$1+2^{50} = 1+4\cdot2^{48} = 1^4 + 4\cdot (2^{12})^4$ Thus by Sophie Germain we got $(2^{25}+2^{13}+1)(2^{25}-2^{13}+1)$

No is prime number, because $2^{50}$ ends in 24, the same digits of $2^{10}$ . Then $2^{50}+1$ has the last digit equal to five. Therefore $2^{50}+1$ does not prime number.

$2^{50}+1 = (2^{2}+1)\times (2^{48}-2^{46}+2^{44}-2^{42} + \cdots - 2^{6} +2^{4}-2^{2}+2^{0})$

through a telescoping series expansion. Hence $2^{50}+1$ is divisible by 5.

2^50 +1= 2^50+2 2^25+1-2 2^25=(2^25+1)^2-2*2^25=(2^25+1)^2-(2^13)^2=(2^25+1+2^13)(2^25+1-2^13)

$2^2 \equiv 4 \equiv -1 \pmod{5}$ . Since 25 is odd, $(-1)^{25} \equiv -1 \pmod{5}$ , so $2^{50} \equiv (2^2)^{25} \equiv -1 \pmod{5}$ . Hence, $2^{50}+1 \equiv 0 \pmod{5}$ , so $2^{50}+1$ is divisible by 5. It follows that $2^{50}+1$ is not prime.

Powers of any number follow a cycle in their last digit. For powers of 2, it's 2, 4, 8, 6 and then repeating. The cycle length is 4. So just as 2^4 end in 6, 2^48 will also end in 6. 2^50, having two more factors of 2, ends in 4. Therefor 2^50+1 ends in 5, which means it must be divisible by 5.

By Jim Schuller: 2*2 + 1 = 5. And 5 multiplied by any number of 2's is going to be divisible by 5. Hence 2^2+1 cannot be prime.

We know that the last digit of $2^{4n+1}$ is 2,the last digit of $2^{4n+2}$ is 4,the last digit of $2^{4n+3}$ is 8 and the last digit of $2^{4n}$ is 6. Since $50=2+12*4$ so the last digit of $2^{50}$ is 4 so the last digit of $2^{50}+1$ is 5 and therefore it divisible by 5 and that means it not prime as well as the answer is No

$(1+p^{10})(1-p^{10}+p^{20}-p^{30}+p^{40}) = 1+p^{50}$ is a non-trivial factorization of $p^{50}+1$ , so it is not a prime number.

$2^{50}+1= 4^{25}+1\equiv(-1)^{25}+1\equiv-1+1=0 \pmod{5}$

It's divisible by 5 so it's not a prime

2^50+1 = (2^10)^5 +1 = ((1024)^4) *1024 +1 = (a number ending in 6) * 1024 + 1 = a number ending in 4 + 1 = number ending 5 = divisible be 5 = composite

Complete the square: 2^50 + 1 = (2^25)^2 - 2 2^25 + 1 - 2 2^25 = (2^25 + 1)^2 - 2*2^25 = (2^25 +1)^2 - 2^26 = (2^25 + 1)^2 - (2^13)^2 Difference of two squares = [(2^25 + 1) + (2^13)] * [(2^25 + 1) - (2^13)] composite

By binomial theorem we can write 2^50 i.e. 4^25 as (5-1)^25 . Last term in this expansion would be -1 which would cancel out the +1 while all other terms will contain atleast one '5' making 2^50 +1 a multiple of 5. Hence it isnt a prime number.

2^10=1,024, hence 2^20 ends by 6 4*4), 2^40 ends by 6, too. Finally, 2^50 ends by 4, and 2^50+1 ends by 5, thus divides by 5

$2^{50}+1=(5-1)^{25}+1≡0 \mod 5$

no here are the factors 1 | 5 | 25 | 41 | 101 | 125 | 205 | 505 | 1025 | 2525 | 4141 | 5125 | 8101 | 12625 | 20705 | 40505 | 103525 | 202525 | 268501 | 332141 | 517625 | 818201 | 1012625 | 1342505 | 1660705 | 4091005 | 6712525 | 8303525 | 11008541 | 20455025 | 27118601 | 33546241 | 33562625 | 41517625 | 55042705 | 102275125 | 135593005 | 167731205 | 275213525 | 677965025 | 838656025 | 1111862641 | 1376067625 | 2175126601 | 3389825125 | 4193280125 | 5559313205 | 10875633005 | 27796566025 | 54378165025 | 89180190641 | 138982830125 | 219687786701 | 271890825125 | 445900953205 | 1098438933505 | 2229504766025 | 5492194667525 | 9007199254741 | 11147523830125 | 27460973337625 | 45035996273705 | 225179981368525 | 1125899906842625 (64 divisors)

It is divisible by 5. Hence not a prime ( $2^{50}+1\equiv0\pmod{5}$ )

2^50 ends with 4 so 2^50+1 ends with 5

2^50 = 4^25. We know that powers of 4 modulo 10 alternate between 4 and 6, so odd powers yield 4 and even powers yield 6. 25 is odd, so 4^25 mod 10 is 5. So, adding 1 gives you 6 mod 10, which is divisible by 2, hence, 2^50 + 1 is not prime.

2^(4x)= AA...AA6

2^((4x)+1)= BB...BB2 (..AA6 *2)

2^((4x)+2)= CC...CC4

(2^((4x)+2))+1=CC...CC5

Where x is a positive integer (12 in this case)

--> CC...CC5 is divisible by 5; end digit is key

The ones digit of the powers of 2 follow the pattern 2,4,8,6. Therefore, 2 to the 48th would go through this cycle 8 times, and 2 to the 50th would have a ones digit of 4. Adding one would make the ones digit 5, and so the number wouldn't be prime because it is divisible by 5 and it is not 5.

The answer can be found using the Wilson's Theorem which states that an integer $p \gt 1$ is a prime iff

$(p-1)!\equiv -1(\mod p)$

If we now substitute $p=2^{50}+1$ and add one to both sides, we get:

$2^{50}!+1\equiv 0 (\mod 2^{50}+1)$

Because ( $2^{50}+1) \nmid (2^{50}!+1)$ , the assumption is wrong and $p$ is not a prime.

2^50 = 4^25. 4 is congruent to 4 mod 5. Any even power of 4 is congruent to 1 mod 5, and any odd power is congruent to 4 mod 5. 4^25 + 1 must be divisible by 5.

The number $2^{50}+1$ is divisible by $5$ (and clearly greater than $5$ ) and therefore cannot be a prime number, since

$2^4 \equiv 1 \mod 5$

and thus

$2^{50} = (2^4)^{12} \times 4 \equiv 4 \mod 5$

whence

$2^{50} + 1 \equiv 4 + 1 \equiv 0 \mod 5$ .

2^(2n+1) is prime for all n is an element of Z^+, and 2^2n is not prime for all n is an element of Z^+

Lo he dividido por todos los anteriores con mi calculadora y había uno que sí.

(2^50)+1=((2^10)^5)+1

Since (2^10)=1024 and,

The result of ((Unit place of 1024 i.e. 4)^5) ends with 4,

Then 4+1=5.

Any natural number ending with 5 necessarily CANNOT BE PRIME.

Hence proved.

This is perhaps the easiest logical solution to the given example!!! Hope it's clearly understood by all.

It's in 2^m +1 form, as per Fermate prime number rule it can be a prime if and only if m is 2^n. In our case 50 cannot be written as 2^n so 2^50 +1 is not a prime.

i just guessed it...and i was right ....after all there are only two choices..what to worry about ..huh

$2^2\equiv -1\pmod{5}$
therefore
$2^{50} \equiv -1\pmod{5}$
also
$1 \equiv 1 \pmod{5}$
adding both
$2^{50} +1 \equiv 0 \pmod{5}$
so not a prime

I've checked the statics of 2^n+1 And it's N y y n y n n n y n n n y...

*n: no. *y: yes

Sol I: Obviously, there are more n then y So, it is not prime

Sol II: Since the pattern begin at the fifth tribe, so (50-4) mod 4 : 2

y n n n y n n n ...

So, it is NOT a prime

Both solutions are showing that it's not a prime number

Nope. Since $2^n\equiv 2\mod 5$ when $n\equiv 1\mod 4$ , then $2^{49}\equiv 2$ , and $2^{50}\equiv 4\mod 5$ , thus $2^{50}+1\equiv 0\mod 5$ so $5$ divides $2^{50}+1$ and so that's not a prime.