4

We have $$(x-1)^3< x^3-2x^2-1 <x^3$$ for all $x\notin \{0,1,2,3\}$ So the only solution for $x$ are in $\{0,1,2,3\}$


4

Find the exponent of the largest power of 2 that divides both sides. In RHS it is $0 + 1 + \ldots + (n-1) = \frac{(n-1)n}{2}$. In LHS it can be found with Legendre's formula, which gives $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor$. Since $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor < \sum_{i=1}^{\infty} \frac{k}{2^i} = k$ (the inequality is ...


3

Via Legendre's formula, we have for any prime number $p$ $$ \nu_p(k!)=\sum_{i=1}^{\infty}\left \lfloor \frac{k}{p^i} \right \rfloor \le \sum_{i=1}^{\infty} \frac{k}{p^i} =\frac{k}{p-1} $$ Note that $\nu_2(2^n-2^i)=i$, and hence $$\nu_2(k!)=\nu_2\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)=\sum_{i=0}^{n-1}i=\frac{n(n-1)}{2}$$ Thus $$\frac{n(n-1)}{2}\le k \tag{...


3

Consider $$(x-1)^3\leq y^3<x^3.$$ The right inequality is true for all integer $x$. The left inequality is true for $x(x-3)\geq0.$


2

Yes, you have $$\sum_k \binom{p}{2k}(-a)^kb^{p-2k}=\pm1.$$ But $b$ is a factor of the LHS, and so of $\pm1$. Thus $b=1$ or $-1$ and all the $b^{p-2k}=b=1$ or $-1$. Using this gives Pollack's formula.


2

Since $p,q,r,s\in\mathbb{N}$, therefore$$p\ge q\ge r\ge s\implies2^p\ge 2^q\ge 2^r\ge 2^s$$Thus, $2^p+2^q+2^r>2^s$, which clearly shows that there are no such $(p,q,r,s)$ that satisfies $$2^p+2^q+2^r=2^s$$


1

HINT. ►The fondamental unit of the field $\mathbb Q(\sqrt{14})$ is easily find (besides of the continued fraction) with the first $b^2$ such that $14b^2$ be an square minus $1$ so we determine from $14\times(4^2)=15^2-1$ that all the solutions of $x^2-14y^2=1$ are given by $$a_n+b_n\sqrt{14}=(15+4\sqrt{14})^n$$ ►If $x_i^2-14y_i^2=1$ and we want to solve $...


1

If $a$ is even, then $$\sum_k\binom{p}{2k}(-a^2)^k\equiv\binom{p}{0}=1\pmod 4.$$ If $a$ is odd, then $$\sum_k\binom{p}{2k}(-a^2)^k\equiv\sum_k\binom{p}{2k} =2^{p-1}\equiv0\pmod 2.$$ So $a$ must be even, and the first case applies.


1

I have always found Burton's number theory book "just right" for a first course in elementary number theory; in addition, he has, in my opinion, an extremely well written, informative, book on the history of mathematics. If you are a more advanced student, you may find Hardy and Wright more agreeable, but if not, Burton's text might be a good place to start....


1

According to Wikipedia, An Introduction to the Theory of Numbers is a classic book in the field of number theory, by G. H. Hardy and E. M. Wright. The fourth edition (1975 printing) is available on line. The updated sixth edition was published in 2008.


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