# Tagged Questions

Questions on congruences, linear diophantine equations, greatest common divisor, divisibility, etc.

100 views

### How can one show the inequality

Let $a,b,n$ be natural numbers (in $\mathbb{N}^*$) such that $a>b$ and $n^2+1=ab$ How can one show that $a-b\geq\sqrt{4n-3}$, and for what values of $n$ equality holds? I tried this: We suppose ...
56 views

### Find the value of p, q, r if $\frac{p}{q + r - p} = \frac{q}{p + r - q} = \frac{r}{p + q - r}$

I equated these 3 to k: $\frac{p}{q + r - p} = \frac{q}{p + r - q} = \frac{r}{p + q - r}$ = k and got k=1. After this, $\frac{p}{q + r - p} = k = 1$, and hence i got: $2p = q + r$ , $2q = p + r$ ...
99 views

### Circular definition?

I am new to logic and studying it by Enderton's A Mathematical Introduction to Logic (2nd edition). For question 3.3.1 in the books, which states Show that in the structure $(\mathbb{N}; \cdot, E)$...
112 views

### Show $a^p \equiv b^p \mod p^2$

I am looking for a hint on this problem: Suppose $a,b\in\mathbb{N}$ such that $\gcd\{ab,p\}=1$ for a prime $p$. Show that if $a^p\equiv b^p \pmod p$, then we have: $$a^p \equiv b^p \pmod {p^2}.$$ ...
63 views

### Primes, congruence mod4, odd exponents, sum of two squares

I am trying to prove: (T) If a prime $p$ is congruent to $3 \bmod 4$ and it occurs with an odd exponent in the prime factorization of $n\in\mathbb N$, then $n$ is not a sum of two squares. I have ...
34 views

62 views

### When is sum of first $n$ natural numbers a square? [duplicate]

My question is: When is $\sum_{i=1}^n i$ a square number? I know that this means i have to solve $n(n+1)/2=m^2$. I tried with modulo 2 etc. but i don't get to it. Please help.
98 views

### $x^4 = -1$ (mod $p$) implies p = 1 mod 8

Let $p$ be an odd prime. Show that $x^4 = -1$ (mod $p$) has a solution if and only if $\Leftrightarrow p = 1$ (mod $8$)
60 views

### How to see that $\text{gcd}(a,b) = \text{gcd}(a-b,b)$?

I'm trying to understand why $\text{gcd}(a,b) = \text{gcd}(a-b,b)$. What is clear to me is that the $\text{gcd}$ divides $a,b$ and also $a-b$ (let's assume $a\ge b$). But then it seems to me we ...
### Let $a,b,m,n \in N$ with $\gcd(m,n)=1$ prove that the modular system $\{ x=a \mod m ; x =b\mod n \}$ has absolution and is unique modulo $mn$}
Let $a,b,m,n \in N$ with $\gcd(m,n)=1$ prove that the modular system $\{ x=a \mod m ; x =b\mod n \}$ has absolution and is unique modulo $mn$} Note that had asked a question I got why there it is ...