Question

Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

score 6.7k · Answer 1 · 2026-04-04 01:31:56

6.7k

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Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

Published on 04 Apr 2026 - 1:31

score 59 · Answer 2 · 2026-04-07 03:57:30

59

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Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$

Published on 07 Apr 2026 - 3:57

score 147 · Answer 3 · 2026-05-10 16:00:21

147

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The Chinese hypothesis revisited

Published on 10 May 2026 - 16:00

score 2.8k · Answer 4 · 2026-05-10 13:50:06

2.8k

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Greatest common divisor of $X^n-1$ and $X^m-1$

Published on 10 May 2026 - 13:50

score 324 · Answer 5 · 2026-04-16 13:05:09

324

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Prove $15\mid 4^{2n}-1$ without induction

Published on 16 Apr 2026 - 13:05

score 75 · Answer 6 · 2026-05-10 14:59:40

75

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Prove: If $ (a^2 \mid (2b^2))$, then $ (a \mid b). $

Published on 10 May 2026 - 14:59

score 77 · Answer 7 · 2026-04-14 00:14:56

77

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A sequence+ divisibility proof contest math

Published on 14 Apr 2026 - 0:14

score 73 · Answer 8 · 2026-05-10 14:07:19

73

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Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Published on 10 May 2026 - 14:07

score 81 · Answer 9 · 2026-04-17 14:44:47

81

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Find $m\in \mathbb{R}$ so that $P(x)$ is divisible by $X^2+X+1$

Published on 17 Apr 2026 - 14:44

score 4.3k · Answer 10 · 2026-04-09 10:26:43

4.3k

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Is $0$ divisible by $0$?

Published on 09 Apr 2026 - 10:26

score 1.8k · Answer 11 · 2026-04-29 12:49:00

1.8k

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Test for divisibility by 2 for a number written in base 9

Published on 29 Apr 2026 - 12:49

score 886 · Answer 12 · 2026-04-15 17:08:21

886

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Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$

Published on 15 Apr 2026 - 17:08

score 397 · Answer 13 · 2026-05-10 15:52:09

397

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The $2n$-digit number divisible by the product of two numbers within itself

Published on 10 May 2026 - 15:52

score 748 · Answer 14 · 2026-04-15 21:59:12

748

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Prove by Induction if x is an odd integer, then for every integer n ≥ 0, $x^{2^n}$ ≡ 1 (mod $2^{n + 1}$)

Published on 15 Apr 2026 - 21:59

score 73 · Answer 15 · 2026-04-30 13:46:11

73

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Sequences and divisibility

Published on 30 Apr 2026 - 13:46

List Question

Prove that the nth prime number $p_n$ satisfies $p_n\leq 2^{2n-1}$

Prove that if $a|b_1b_2$, $a|c_1c_2$ and $a|b_1c_1+b_2c_2$, then $a|b_1c_1$ and $a|b_2c_2$

The Chinese hypothesis revisited

Greatest common divisor of $X^n-1$ and $X^m-1$

Prove $15\mid 4^{2n}-1$ without induction

Prove: If $ (a^2 \mid (2b^2))$, then $ (a \mid b). $

A sequence+ divisibility proof contest math

Find divisibility for $n \in \mathbb {Z+}, 2\nmid n, 2^n-1, 2^n +1$ w.r.t. $5$ using congruence arithmetic.

Find $m\in \mathbb{R}$ so that $P(x)$ is divisible by $X^2+X+1$

Is $0$ divisible by $0$?

Test for divisibility by 2 for a number written in base 9

Prove that $76$ raised to any integer power and then divided by $100$ (ignoring the remainder) is always divisible by $57$

The $2n$-digit number divisible by the product of two numbers within itself

Prove by Induction if x is an odd integer, then for every integer n ≥ 0, $x^{2^n}$ ≡ 1 (mod $2^{n + 1}$)

Sequences and divisibility

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