# Tagged Questions

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### Indcution and Greatest Element in a Subset

I wish to show that the following two assertions are equivalent: (Principle of Mathematical Induction) Let $S$ be a nonempty subset of the set of non-negative integers satisfying the following two ...
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### Is it true that $n^2+3n+13$ is prime for all $n\in\mathbb ℤ^+$?

Prove or disprove the statement: If $n\in\mathbb ℤ^+$, then $n^2+3n+13$ is prime. I am lost here. All I know is that $n$ is greater than or equal to one, since it is a positive integer.
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### Proving that $3 = 9^{-1} \pmod{26}$

Prove that $3$ is the multiplicative inverse of $9 \pmod {26}$ $$\quad26\quad1\quad0\\2\quad9\quad0\quad1\\\;\;1\quad8\quad1\quad{-2}\\\quad\;1\quad-1\quad3$$ Hence $3$ is the multiplicative inverse ...
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### $\gcd(a,b)$ compared to $\gcd(3a,b)$

$\gcd(a,b)=\gcd(3a,b)$? They are obviously not equal in general, as $\gcd(ax, bx)=|x|\gcd(a,b)$.
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### Pair of positive integers in product sums

I am still not sure on this answer. I would like someone to help me see the solution to his question. I was working on it for a while and it is the only question that I looked at that I can not ...
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### Using jugs filled with water problem

Given jugs $m$ and $n$ liters (WLOG $m<n$) is it always possible to get all $i$, $0 \leq i \leq n ?$ If so, prove it. If not, explain which $i$ you can get. Is there also a minimum number ...
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### Prove for all whole numbers n, $(n+1)(n+2)…(2n-1)(2n)=2^n(1)(3)(5)…(2n-1)$ [duplicate]

Prove for all whole numbers n, $(n+1)(n+2)...(2n-1)(2n)=2^n(1)(3)(5)...(2n-1)$. I got upto $(2n)(2n-2)(2n-4)...=2^n$, after which I'm stuck.
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### Seeking help extending Vieta-jumping to higher powers

I am trying to prove the following conjecture. Conjecture. If $r > s \ge 1$ are relatively prime integers such that $$(r-s)^4-1 \equiv 0\!\pmod{4r^2s}, \tag{1}$$ ...
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### Prove that $133$ divides $11^{n+1} + 12^{2n−1}$ for all $n > 1$? [duplicate]

I have tried this question so hard but still stuck here. It seems like easily provable if all $n$ are all positive numbers but in this question, the $n$ is bigger than $1$. original question : prove ...
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### Proving a set of numbers has arithmetic progressions of arbitrary length, but none infinite

For each real number $x$, let $[x]$ be the largest integer less than or equal to $x$. For example, $$[5] = 5$$ $$[7.9] = 7,$$ and $$[−2.4] = −3.$$ An arithmetic progression of length $k$ is a ...
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### Prove $x^n-1=(x-1)(x^{n-1}+x^{n-2}+…+x+1)$

So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$ I think that to prove this I should use induction, however I am a bit stuck ...
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### If $n$ is any positive integer whose last digit is $5$, then $5$ divides $n$

Prove that if n is any positive integer whose last digit is a 5, then 5|n Therefore, n is going to be 5, 15, 25, 35 etc ... b∣a states that 'b divides a' and we know that 5∣5, 5∣15, 5∣25, 5∣35 ...
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### Ground Plan - Prove Fermat-Euclid's Totient Theorem with Lagrange's Theorem

If $\gcd(a,n) = 1$, then $a^{\phi(n)}\equiv 1\pmod n$. Here's a three-step proof. An integer a is invertible means there's some $a^{-1}$ such that $aa^{-1}\equiv 1 \pmod n$. By cause of Jones p84 ...
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### Fermat's Little Theorem fails for composite instead of prime numbers.

I know Fermat's Little Theorem = Fermat-Euler's Totient Theorem when $n$ is prime. Elementary Number Theory, Jones, p83 writes if we simply replace p with a composite integer n, then the ...
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### Backward direction – Wilson’s Theorem – p is prime $\iff (p-1)!\equiv-1(mod\ p)$.

(1) How can you preconceive to prove by contradiction? Prove by contradiction. Suppose $n$ is composite. This means there exists a divisor $d|n$ such that $1<d<n$. We are given that ...
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### Ground Plan – Forward direction – Wilson’s Theorem – p is prime $\iff (p-1)!\equiv-1(mod\ p)$.

Lemma 5.3 - I omit proof here - Let p be prime. Then $x^2 \equiv 1 \, (mod p) \iff x \equiv \pm 1 \; (mod p)$ First we establish the result for the first two primes 2, 3. Then prove the result for ...
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### Ground plan of Forward direction - Let $p$ be an odd prime. Prove $x^{2} \equiv -1 \; (mod \, p)$ has a solution $\iff p\equiv 1 \; (mod 4)$

Prove by contradiction. Thence suppose NOT $p\equiv 1 \; (mod 4)$. Thence 3 possibilities remain: $4|p, 4|(p - 2), 4|(p - 3)$. But $p > 2$ is prime, thence $4 \not | p$. (1) How can you ...
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