I have seen the past threads but I think I have another proof, though am not entirely convinced. Suppose there are only finitely many primes $p_1, …, p_n$ of the form $6k-1$ and then consider the number $N = (p_1.p_2. … .p_n)^2 – 1$ If all of the primes dividing N are of the form […]

Now a days, I become good fan of this site, as this site making me to learn more math..hahaha. Okay! Can we prove that $x^3 + 7$ cannot be perfect square for any positive/negative or odd/even integer of $x$. I checked with number up to x = 1,…1000. I realized that, not a perfect. But, […]

If $n$ is a positive integer, let $\phi(n)$ the Euler function. ( if $n=p_1^{\alpha_1}\dots p_k^{\alpha_k}$ with $p_i$ distinct primes, we have $\phi(n)=p_1^{\alpha_1-1} \dots p_k^{\alpha_k-1}(p_1-1)\dots(p_k-1)$ ) Let $P$ a polynomial in $\mathbb{Z}[X,Y]$. We suppose there exists an infinite number of positive integer $n$ such that $P(n,\phi(n))=0$ Is $P$ reducible in factors of degree one ? Thanks […]

Prove by infinite descent that there do not exist integers $a,b,c$ pairwise coprime such that $a^4-b^4=c^2$.

Let $A:=\sum_{k=0}^{\infty}x^{2^k}$. For what $n$ is it true that $(A+1)^n+A^n\equiv1\mod2$ (here we are basically working in $\mathbb{F}_2$.) The answer is all powers of 2, and it’s fairly simple to see why they work, but the hard part is proving all non-powers of 2 don’t work. In the solution, it says “If $n$ is not a […]

Each of the numbers $a_1 ,a_2,\dots,a_n$ is $1$ or $−1$, and we have $$S=a_1a_2a_3a_4+a_2a_3a_4a_5 +\dots+ a_na_1a_2a_3=0$$ Prove that $4 \mid n$. If we replace any $a_i$ by $−a_i$ , then $S$ does not change $\mod\, 4$ since four cyclically adjacent terms change their sign. Indeed, if two of these terms are positive and two negative, […]

How to find necessary and sufficient conditions for the sum of two numbers to divide their product. Thanks in advance.

Define $\sigma(i)$ to be the sum of all the divisors of $i$. For example, $σ(24) = 1+2+3+4+6+8+12+24 = 60$. Given an integer $n$, how can we count the number of integers $i$, less than or equal to $n$, such that $\sigma(i)$ is even?

Let $m$ be odd and let $a \in \mathbb{Z}.$ The congruence $x^2 \equiv a \mod m$ has a solution if and only if for each prime $p$ dividing $m,$ one of the following conditions holds, where $p^{\alpha} \mid \mid m$ and $p^{\beta}\mid \mid a$: $\beta \geqslant \alpha$; $\beta < \alpha$, $\beta$ is even, and $a/p^{\beta}$ […]

This question already has an answer here: Showing that $a^n – 1 \mid a^m – 1 \iff n \mid m$ 3 answers

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