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16 mar 2017 · Factorize $n^2-1$ into $(n-1)(n+1)$. If $n$ is even, both terms are odd. If $n$ is odd, one term is $2\pmod4$ and the other $0\pmod4$ hence it is divisible by $8$.
- 4^n+ 1
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- 4^n+ 1
27 lip 2018 · Notice that if $n^2-1$ is divisible by $8$, then $n^2-1=8k$ for some integer $k$. Then $n^2=8k+1=2(4k)+1$. Letting $4k=j$ gives us $n^2=2j+1$ which is an odd integer.
5 lis 2021 · As an integer is Since 2 and 4 are common factors of $4k^2 + 4k$, which means 8 must also be a factor of $4k^2 + 4k$. $n^2-1$ is therefore divisible by 8 where $n = 2k+1$ for some integer k. Therefore, we have proven that $n^2-1$ is divisible by 8 for any odd integers n.
4 lip 2018 · Since #(2n+1)^2-1# is divisible by 8 whenever #(2n-1)^2-1# is divisible by 8 and #(1)^2-1# is divisible by 8, we have shown that #a^2-1# is divisible by 8 for all odd #a#.
Let \(x\) be a real number. Prove by contrapositive: if \(x\) is irrational, then \(\sqrt{x}\) is irrational. Apply this result to show that \(\sqrt[4]{2}\) is irrational, using the assumption that \(\sqrt{2}\) is irrational.
N k= −2 1 and k = −2 1p, p∈ . Use direct proof to show that 2p+1 is a factor of N. SYN-K , proof
17 kwi 2022 · That is, prove that if \(r\) is a real number such that \(r^3 = 2\), then \(r\) is an irrational number. Prove the following propositions: (a) For all real numbers \(x\) and \(y\), if \(x\) is rational and \(y\) is irrational, then \(x + y\) is irrational.
