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$n^2 = 1 \pmod{24}$ for $n=1,5,7,11$, by checking each case individually. $(n+12)^2 = n^2 + 24n + 144 = n^2 \pmod{24}$. Therefore, $n^2 = 1 \pmod{24}$ when $n$ is odd and not divisible by $3$, and so $n^2-1$ is divisible by $24$ for these $n$. You don't need primality of $p$ here!
17 Απρ 2022 · First, multiply both sides of the inequality by xy, which is a positive real number since x > 0 and y > 0. Then, subtract 2xy from both sides of this inequality and finally, factor the left side of the resulting inequality. Explain why the last inequality you obtained leads to a contradiction.
18 Φεβ 2021 · Prove or disprove: \(2^n+1\) is prime for all nonnegative integer \(n\). Solution. Consider \(n=3;\) \(n\) is a nonnegative integer. \[2^n+1=2^3+1=9.\] \(9\) is not prime, since (\(3)(3)=9\), thus the statement: \(2^n+1\) is prime for all nonnegative integer \(n\) is false.
26 Απρ 2023 · Theorem. Let a ∈ N a ∈ N be an even perfect number . Then a a is in the form: 2n−1(2n − 1) 2 n − 1 ( 2 n − 1) where 2n − 1 2 n − 1 is prime . Similarly, if 2n − 1 2 n − 1 is prime, then 2n−1(2n − 1) 2 n − 1 ( 2 n − 1) is perfect .
13 Ιουν 2020 · By the distributive law, $(x + 1)(x - 1) = x^2 - 1$. Set $x = -1$ and observe that the left hand side is zero, so the right hand side is zero, so $x^2 = 1$. It's arguable whether the assumptions I use there are any more or less obvious than the conclusion.
26 Φεβ 2015 · Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. My work: So I think I have to do a proof by induction and I just wanted some help editing my proof. My attempt: Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$ $$1=\frac{6}{6}.$$ So $P(1 ...
x^2: x^{\msquare} \log_{\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int_{\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x)
