Problem is as stated in the title. Source is Larson’s ‘Problem Solving through Problems’. I’ve tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing’s clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is […]

How can I show that $(n-1)!$ is congruent to $-1 \pmod{n}$ if and only if $n$ is prime? Thanks.

Well I thought this is obvious. since $\gcd(a,b)=1$, then we have that $a$ does not divide $b$ AND $a$ divides $bc$. this implies that $a$ divides $c$. done. but apparently this is wrong. help explain why this way is wrong please. the question tells you give me two relatively prime numbers $a$ and $b$ such […]

As the title stated, I am wondering the integers $x,y$ that satisfy the equation $x^2-y^3 = 23$.

I was given this proposition but I was never able to prove it. Does anyone know how to solve this? Let $f$ be a polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$, where $p$ is prime. Then $f$ has at most $\deg f$ roots.

Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$. I tried hard but nothing seems to work. 🙁

Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions. My attempt: We have, $x^2 – 1 = (x – 1) \times (x + 1)$, then $(x – 1)(x + 1) \equiv 0 \pmod{2^k}$ which implies, $2^k|(x – 1)$ or $2^k|(x + 1) \implies x \equiv \pm 1 \pmod{2^k} (1)$ Furthermore, $2^{k-1} \equiv 0 […]

Let $a,b,c \in \mathbb N$ find integer in the form: $$I=\frac{a}{b+c} + \frac{b}{c+a} + \frac{c} {a+b}$$ Using Nesbitt’s inequality: $I \ge \frac 32$ I am trying to prove $I \le 2$ to implies there $\nexists \ a,b,c$ such that $I\in \mathbb Z$: $$I\le 2 \\ \iff c{a}^{2}+3\,acb+{a}^{3}+{a}^{2}b+a{b}^{2}+{b}^{3}+c{b}^{2}+{c}^{2}b+{c }^{3}+{c}^{2}a-2\, \left( b+c \right) \left( c+a \right) \left( […]

I’m trying to prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$ I showed that both $n,m$ divides $nm/\gcd(n,m)$ but I can’t prove that it is the smallest number. Any help will be appreciated.

Show that $13$ divides $2^{70} + 3^{70}$. My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat’s little theorem. So how can I separate the two? Thanks!

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