I tried to use quadratic reciprocity, but I don’t understand how to use the explicit formulae to end the problem. I surmised that the primes that work are congruent to $\pm1\bmod5$, but I could not generalise. where $F_{p-1}$ is the p-1th term of the Fibonacci Sequence

The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4).$ Find 3 more solutions in positige integers. [hint. look for solutions of the form $(a,b,c) = (xz, yz, z^2)$ Attempt: So I tried to use the hint in relation to the triple that they gave that worked. So I observed that […]

I am looking for a systematic way of deciding if a given number is a square in $\Bbb Z/n\Bbb Z$. E.g. is $89$ a square in $\Bbb Z/n\Bbb Z$ for $n\in \{25,33,49\}$? Brute-forcing it would take too long here. How else can we do this?

For a problem set, we had two find the final two digits of 17^17^17 So what I did was find the last digit of 17^17 and then take 17^of that last digit of 17 and then find the last two digits of that number. I got the last two digits as 17. Is my method […]

The number $198$ has a curious property. If we factorise it as $198 = 2\times3^2\times11$ and factorise its “reverse” as $891 = 3^4\times11$ and sum up the numbers which appear in each factorisation: $$ 2+3+2+11 = 18 \quad\text{ and }\quad 3+4+11 = 18\;, $$ we obtain the same number. Indeed, all of $18, 198, 1998, […]

During class, we were introduced to a proof that used the ceiling function. We assumed (without proof) that: $$ \left\lceil{\frac{n}{2^i}}\right \rceil= \left\lceil{\frac{\lceil{\frac{n}{2}}\rceil}{2^{i-1}}}\right\rceil,$$ for $i \geq 1$, where $d$ is a positive integer. I am interested in seeing how this can be proved. I don’t know much about the ceiling function, other than its basic properties. […]

I’m trying to prove that for every integer $n$, $15\mid n$ iff $3\mid n$ and $5\mid n$. The first part of this bi-conditional was easy for me to prove, but I’m having problems with the second. Here is what I have so far: Suppose $15\mid n$. Then we can let $k$ be some integer such […]

The question goes like this: use the CRT to prove that if an integer $n>1$ is not a power of a prime, then there exists an integer $x$ such that $n|(x^{2}-x)$ but $n$ does not divide $x$ nor $x-1$. I can see this working with a simple example like $n=12$, which lets $x=4,-3$. I got […]

When I graphed the relation $a^{a+2b}=b^{b+2a}$ , it gives a graph similar to $y=x$. However, the question explicitly states that $a \neq b$. So does that mean that no such $N$ exists ? What happens when the problem is generalized as $N=a^{ma+nb}=b^{mb+na} $ ? Can anybody help as to what should be done ? Thanks […]

Argue that there are infinitely many primes p that are not congruent to 1 modulo 5. I find this confusing. Is this saying $p_n \not\equiv 1 \pmod{5}$? To start off I tried some examples. $3 \not\equiv 1 \pmod{5}$ $5 \not\equiv 1 \pmod{5}$ $7 \not\equiv 1 \pmod{5}$ $11 \equiv 1 \pmod{5}$ $13 \not\equiv 1 \pmod{5}$ $17 […]

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