let $n$ be a natural number and $r$ , $s$ be rational such that   $n=s^2+r^2$  show that there are natural numbers a,b such that   $n=a^2+b^2$
46398
Let $a\in \mathbb Z$. Is it true that $a$ is a sum of two squares of rational numbers if and only if it is a sum of two squares of integers?  I came to face this problem while dealing with quaternion algebras.
58156
Let $a$ and $b$ be positive real numbers and let $n$ be a natural number prove that   $$\left({1+\frac ab}\right)^n+\left(1+\frac ba\right)^n\ge2^{n+1}.$$
8721
Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:  $$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$  for some integers $m, n$  $$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$  $$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$  Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.   For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.  $$\implies (p + q)^2 = \frac{m^2}{n^2}$$  $$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$  Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.  Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.   $$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$  is rational as well.  So for all $i$,  $$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$  and   $$a^{\frac{i}{3}}b^{\frac{i}{3}}$$  are rational.  I know I'm really close to the answer, but it somehow just keeps slipping through my fingers.
90597
Let $a&gt;0$ be a real number, such that for all integers $n\geq 1$: $n^a \in \mathbb N$ Show that $a$ must be an integer.    It's not difficult to show this when $a$ is a rational number: $2^\frac{p}{q}$ is irrational when the fraction is in lowest terms and $q \neq 1$.  When $a$ is irrational, then for all $n\geq 1$, there exists $m_n\in \mathbb N^*$, such that: $$a = \frac{\log m_n}{\log n}; \quad \text{$m_n$ is not a power of $n$}$$  I think considering $n=2,3$ is enough to show a contradiction, but I can't seem to find it. This is what I get: $$a = \frac{\log p}{\log 2} = \frac{\log q}{\log 3}$$ $$p = 2^{\log q/\log 3} $$  I think the RHS is irrational when $q$ is not a power of $3$, but I can't prove it. The closest thing I have to a solution is . But it uses an unproven conjecture, and I was hoping for a more elementary proof.
96833
How does one prove by induction that  $n! &gt; n^2$  for $n \geq 4$
58486
Show that every even natural number grater than $2$ can be expressed as a sum of two prime numbers. No idea how to prove this. Can you help? thanks
75422
I need help with this question.     Let $a$ and $b$ be positive integers such that $a^{n}+n$ divides $b^{n}+n$ for every natural number $n$. Show that $a=b$.   Any help would be appreciated! Thanks!
95872
Show that if real numbers $a_1,a_2,\ldots,a_n$ satisfy  $$a_1^l+a_2^l+\cdots+a_n^l=0$$ for every odd $l$, then for any $a_i$ we can always find some $a_j$ (not necessarily different) such that $a_i+a_j=0$.
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
69440
Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ?
49324
Suppose $A$ is the set of all rational numbers $p$ such that $p^2 &lt;2$ and $B$ is the set of all rational numbers $p$ such that $p^2 &gt; 2$. We want to show that $A$ contains no largest element and $B$ contains no smallest element.  In Rudin's Principles of Mathematical Analysis, he associates $q = p- \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}$. Did he just come up with this using trial and error?
56960
Assume that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational，prove $\sqrt{a},\sqrt{b},\sqrt{c}  \in\mathbb Q$,are rational.  I know that can be proved, would like to know that there is no easier way    $\sqrt a + \sqrt b + \sqrt c = p \in \mathbb Q$, $\sqrt a + \sqrt b  = p- \sqrt c$, $a+b+2\sqrt a \sqrt b = p^2+c-2p\sqrt c$, $2\sqrt a\sqrt b=p^2+c-a-b-2p\sqrt c$, $4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)\sqrt c$, $\sqrt c=\frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}\in\mathbb Q$.
9905
Let $n \in \mathbb N$ be a natural number and $a \in \mathbb R$ be a real number. The $n$-th root of the number $a$ is defined as follows:  Case I: $n$ is an odd number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \in \mathbb R$ such that $b^n = a$.  Case II: $n$ is an even number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \geq 0$ such that $b^n = a$.  Why is it that when $n$ is even, we only consider $b \geq 0$. For example, both $+2$ and $-2$ squared equal $4$, but when we say the square root of $4$ is $2$. Is there a reason for this?
93175
Show that if $r$ is an nth root of $1$ and $r\ne1$, then $1 + r + r^2 + ... + r^{n-1} = 0$.  I think I can represent all the roots of 1 as follows:  $r = 1^{\frac{1}{n}} ( \frac{\cos{2\pi k}}{n} +  i\frac{\sin{2\pi k}}{n} )$  From there I am not sure how to get to $1 + r + r^2 + ... + r^{n-1} = 0$.
56095
If $k, m, n$, are natural numbers and $k \leq n$ What is:  $$\sum_{r=0}^{m}\frac{k\binom{m}{r}\binom{n}{k}}{(r+k)\binom{m+n}{r+k}}$$
38448
Let $x_1, x_2 \in \mathbb R$ and let $(a_0, a_1, \dots, a_n), (b_0, b_1, \dots, b_n)$ be $(n+1)$-tuples of real numbers. Show that there is a unique polynomial of degree at most $2n+1$ such that $$q^{[k]}(x_1)=a_k,$$$$q^{[k]}(x_2)=b_k$$ for $k=0, \dots, n$.   Any hints on how to get started with this exercise?
60388
$x$ and $y$ are real numbers.  Given that $1&lt;x^2-xy+y^2&lt;2$, how can I show that $\frac 29&lt;x^4+y^4&lt;8$ ?  Then can I use that to prove that for any natural number $n&gt;3$  $$x^{2^n}+y^{2^n}&gt;\frac 2{3^{2^n}} \text{?}$$