prime factors of number with a particular form

Question

I try to factorize this huge number $2^{3^{5^{7}}}$+ $7^{5^{3^{2}}}$ .but i have no idea,the only thing i know is that it's not divisible by 7 and 11. can you help me find some prime factors of this huge number ? (it turned out that this number is indeed divisible by 3. thank you for that information.).This number is larger than i thought. it has 0.3A digits,where A itself is a 37276 digit number(!!!!). What about $2^{3^{5^{7^{11}}}}$+$11^{7^{5^{3^{2}}}}$ ?wanna try?

Answer 1

Summary: The number is divisible by 3, by 5, and by 144,883,729. It is not divisible by any other primes less than 1 trillion, nor by higher powers of those three primes. Its last four digits are 5015.

---

It's clear that the number is not divisible by 2 or by 7. In fact, it's equivalent to 1 mod 2 (i.e. it's odd), and to 1 mod 7 (i.e. the number is a multiple of 7, plus 1.) We can see the latter by observing that the powers of 2 form a cycle of length 3, mod 7: $$ 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, \dotsc \equiv 2, 4, 1, 2, 4, 1, \dotsc \pmod 7. $$ Since we're raising 2 to a multiple of 3, the result will be equivalent to 1, mod 7.

Similar methods let us test other primes. As Lucian said, the number is divisible by 3. It is not divisible by 9: the powers of 2 form a cycle of length 6, mod 9: $$ \langle 2^k \rangle_{k \geq 1} \equiv \langle 2, 4, 8, 7, 5, 1, \dotsc \rangle \pmod 9. $$ Since we're raising 2 to an odd multiple of 3, the term $2^{3^{5^7}}$ is equivalent to 8, mod 9. Meanwhile, the power of 7 form a cycle of length 3, mod 9: $$ \langle 7^k \rangle_{k \geq 1} \equiv \langle 7, 4, 1, 7, 4, 1 \dotsc \rangle \pmod 9. $$ So to get $7^{5^{3^2}}$ mod 9, we just need to know $5^{3^2}$ mod 3. Five has order 2, mod 3: $$ 5^1, 5^2, 5^3, 5^4, \dotsc \equiv 2, 1, 2, 1, \dotsc \pmod 3. $$ 5 to an odd power is equivalent to 2. So $7^{5^{3^2}} \equiv 7^2 \equiv 4 \mod 9$. Thus, the sum is equivalent to $8 + 4 \equiv 3 \pmod 9$.

Similarly, it is divisible by 5, but not by 25. Two has order 20, mod 25. Three has order 4, mod 20. And any power of five is equivalent to one, mod 4. So $2^{3^{5^7}} \equiv 2^{3^1} \equiv 2^3 \equiv 8 \pmod{25}$. Seven has order 4, mod 25. And again, any power of five is equivalent to one, mod 4, so $7^{5^{3^2}} \equiv 7 \pmod{25}$. Thus, the sum is equivalent to $15 \pmod{25}$.

With the exact same methods, it's easy to check by hand that the number is not divisible by other small primes. I automated the process with a program in SAGE. See the notebook here. It shows you the remainder modulo each prime up to 1009.

The notebook also uses the Chinese Remainder Theorem and the residues computed for all those primes to find a gigantic 420-digit number. Let's call it $X$. Our number is equal to $X$ plus a multiple of the product of all the primes up to 1009, except we use 9 in place of 3 and 25 in place of 5. Hence $$ 2^{3^{5^7}} + 7^{5^{3^2}} = X + 15 k \prod_{p = 2}^{1009} p $$ for some integer $k$, where the product is over primes.

The notebook also uses the same ideas to find the last three digits of the result, which are $015$. You can do the same thing to find more digits. But since 2 is not a unit mod $10^k$, you can't use the method multiplicative_order(); you have to find the cycle length of its powers manually.

Finally, the notebook checks primes up to 1 billion to find any other factors. The only factor found is 144,883,729. I let it run for a week on my home computer and checked up to 1 trillion: there aren't any more factors.

Answer 2

The only thing I know is that it's not divisible by $3$

$2\equiv-1\bmod3,\quad7\equiv1\bmod3,\quad3^{5^7}$ and $5^{3^2}$ are odd numbers, therefore $(-1)^\text{odd}+1^\text{odd}=0$, hence the number is a multiple of $3$.