query: Prove that $2^{2^5} +1 \equiv 0 \mod 641$
positive_passages.docid: 8364
positive_passages.text: To show that Fermat number $F_{5}$ is divisible by $641$.
negative_passages.docid: 27063
negative_passages.text: Prove that if n is an odd positive integer , then $n^2\equiv1\mod8$
negative_passages.docid: 32803
negative_passages.text: Prove that if b is coprime to 6 then $b^2 \equiv 1 $ (mod 24)
negative_passages.docid: 110752
negative_passages.text: $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$
negative_passages.docid: 105442
negative_passages.text: Show that any prime divisor of $x^4+x^3+x^2+x+1$, with $x\in\mathbb{N}$, is $5$ or $1$ mod $5$
negative_passages.docid: 48875
negative_passages.text: $n^3 \equiv n^5 \pmod{12} $?
negative_passages.docid: 26478
negative_passages.text: Show that the congruence $x^{p-1} \equiv 1 \pmod {p^2}$ has precisely $p-1$ solutions modulo $p^2$
negative_passages.docid: 125868
negative_passages.text: Prove $a \equiv b \pmod{c} \implies a^n \equiv b^n \pmod{c}$.
negative_passages.docid: 5043
negative_passages.text: Show that $2^n>1+n\sqrt{2^{n-1}}$
negative_passages.docid: 11989
negative_passages.text: $ac\equiv bc \pmod{\!m}\!\!\iff\!\! a\equiv b \pmod{\!\!\frac{m}{(c,m)}}$ [Euclid's Lemma, congruence form]
negative_passages.docid: 5708
negative_passages.text: Prove that if $k\ge 1\,$ and $\,a^k\equiv 1\pmod{\! n}$ then $\gcd(a,n)=1$
negative_passages.docid: 132587
negative_passages.text: Prove that the product of coprimes must be congruent to $-1$ or $1 \pmod{m}$
negative_passages.docid: 6160
negative_passages.text: How to prove a number $a$ is quadratic residue modulo $n$?
negative_passages.docid: 141909
negative_passages.text: Prove ${2}^{n}\cdot(2m+1)-1 $ is invertible
negative_passages.docid: 34323
negative_passages.text: Prove that $a^{(p-1)/2} \equiv 1$ (mod p) and $a^{(p-1)/2} \equiv -1 \pmod p$
negative_passages.docid: 119436
negative_passages.text: Proving $2^n\leq 2^{n+1}-2^{n-1}-1$ for all $n\geq 1$ by induction
negative_passages.docid: 111505
negative_passages.text: Proof $\binom{p-1}{k}\equiv (-1)^k \mod{p}$