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5

Let $a=qb+r$ with Euclidian division, $0 \le r< b$, then $$2^a+1 \equiv 2^{qb+r}+1\equiv 2^r+1 \mod 2^b-1.$$ So it can't be congruent to zero. (Moreover, For $b=2$, $a=r=1$ satisfies. The reason is the fact of $3$: $\;\;$ $3=2^2-1=2+1$).


4

You can't reason from $256\mid ab$ to $256\mid a \lor 256\mid b$ because $256$ is not a prime. Instead you need to consider all the ways that the prime factors of $256$ can be distributed among $a$ and $b$. Since $256$ is a prime power, namely $2^8$, there are only nine ways to write it as a product, and we can check those systematically: Case 0. ...


3

Suppose $a = bq + r$, where $0 \leq r < b, q \geq 1$. Assume: $2^a+1 = k(2^b-1) \Leftrightarrow k+1 = k2^b-2^a = 2^b(k-2^{a-b})$. Thus, $k$ is odd, say $k = 2k_1-1$, so $2k_1 = 2^b(2k_1-1-2^{a-b})$. This implies $k_1 =2^{b-1}k_2$. So $k_2 = 2^bk_2-1-2^{a-b}$, or $1+2^{a-b} = k_2(2^b-1)$. Continue this procedure $q$ times, we obtain: $1+2^r=k_q(2^b-1)$. ...


3

You can, in fact, do it with $b=1$, noting that this amounts to looking for a non-square integer of the form $$d={(ap-1)(ap+1)\over q^2}$$ To find such a $d$, note that $\gcd(p,q)=1$ implies there is a number $a_0$ such that $a_0p\equiv1$ mod $q^2$. If we now take $a$ of the form $a_0+kq^2$, we see that $ap+1=(a_0p+1)+kpq^2$, which means, by Dirichlet's ...


2

The ideia is reduce to your first case. Divide $a$ for $b$ and write $a = bq + r$, where $0 \le r < b$. Now you see that $2^a + 1 = 2^{bq+ r} + 1 = 2^r\cdot (2^b)^q + 1$ Use Newton binom and write $(2^b)^q = \sum_{j=0}^{q} \binom{q}{j} (2^b - 1)^j$ Hence, you have: $2^a + 1 = 2^r\{ \sum_{j=0}^{q} \binom{q}{j} (2^b - 1)^j\} + 1$ Now, for $j > ...


2

If $(n+1)^2-1$ is even, then $(n+1)^2$ is odd. But if the square of something is odd, then the something is odd. (Try to prove that) Hence $n+1$ is odd and $n$ is even.


1

$(n+1)^2-1 = n^2+2n+1-1=n^2+2n=n(n+2)$. $n$ and $n+2$ are either both even or both odd. The product of two odd numbers returns an odd number, while the product of two even numbers returns an even number. Hence if $n$ is even, then $n(n+2)$ is even, then $(n+1)^2-1$ is even, and vice versa.


1

Using the difference of squares factorization method we have: $$(n+1)^2-1=(n+1-1)(n+1+1)=n(n+2)=n^2+2n$$ Since $(n+1)^2-1$ is even, we can say $(n+1)^2-1=2a$ for suitable $a\in\Bbb N$. So we obtain $$n^2+2n=2a \Rightarrow n^2=2(a-n)$$ Thus, we see that $n^2$ is an even number (it is divisible by $2$), and so $n$ must also be even.


1

Not an answer, but just to share if it could be a way, on the track of your edit2. Put $z = a + i\,b\quad w = c + i\,d$ then: $$ a^{\,2} + b^{\,2} - c^{\,2} - d^{\,2} = \tilde z\,z - \tilde w\,w $$ $$ \left( {\tilde z + \,\tilde w} \right)\left( {z - w} \right) = \tilde z\,z - \tilde z\,w + \tilde w\,z - \tilde w\,w = \tilde z\,z - \tilde w\,w - ...


1

In the 6th line of the text, $m$ is defined in such a way that $p_r|m$. Hence $\frac{m}{p_r} \in \mathbb{Z}$.


1

You know that $$(\Bbb Z/256\Bbb Z)^\times\cong \Bbb Z/2\Bbb Z \times \Bbb Z/64\Bbb Z$$ and Hensel's lemma gives a lifting for solutions if you can find them modulo $8$ and unique lifts. Since there are $4$ solutions modulo $8$, these lift to $4$ solutions modulo any higher power of $2$. Edit: Since the op doesn't know Hensel, it's easy enough to just ...


1

Differentiating with respect to $x$ gives $$10\ (1+3x+x^2)^{9}(2x+3)=\sum_{r=0}^{20}ra_r x^{r-1}\ $$ Multiply this by $3$ and add to the original expression to get $$30\ (1+3x+x^2)^{9}(2x+3)+(1+3x+x^2)^{10}=\sum_{r=0}^{20}3ra_r x^{r-1}+\sum_{r=0}^{20}a_r x^r$$ $$=\sum_{r=0}^{20}a_rx^{r-1}(3r+x)$$ Since $x$ is an arbitrary number, its value doesn't matter. ...



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