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9

Another way of saying that $f(x)|f(x+f(x))$ is that $f(x+f(x))$ belongs to the ideal $(f(x))$ generated by $f(x)$ in $K[x]$. At the same time this is equivalent to $\pi(f(x+f(x))=\overline{f(x+f(x))}=0$, where $\pi$ is the natural quotient homomorphism $$\pi:K[x]\to K[x]/(f(x)).$$ Now you're done cause $$\overline{f(x+f(x))}=f(\overline{x}+\overline{f(x)})=f(...


6

Note that $Q(x) = (x-1)^2$, which means that $(x-1) \mid P(x)$ and $(x-1) \mid P'(x)$. In other words $x=1$ is a zero of both $P(x)$ and $P'(x)$. So after all you're left to solve the system of linear equation: $$\begin{cases} a-b+1 = 0 \\ 2014a - 2015b = 0 \end{cases}$$ It's easy to conclude that $a=-2015$ and $b=-2014$


6

I'll show that $2^{n+1}\mid\mid T_{2^n}$ for all $n\ge 5$. Together with casework for $n<5$ this gives a proof of your claim. Lemma: For all integers $n,m\ge 1$ we have $$T_{n+m}=T_{m}T_{n-1}+T_{m-1}T_{n}+T_{m}T_{n}+T_{m+1}T_{n+1}.$$ Proof: Straightforward induction on $m$. $\square$ Proposition: For all integers $n\ge 5$: $$T_{2^n}\equiv 2^{n+1}\pmod{...


5

Result of exhaustive search with software: 123648, 123864, 123984, 124368, 126384, 129384, 132648, 132864, 132984, 134928, 136248, 136824, 138264, 138624, 139248, 139824, 142368, 143928, 146328, 146832, 148392, 148632, 149328, 149832, 162384, 163248, 163824, 164328, 164832, 167328, 167832, 168432, 172368, 183264, 183624, 184392, 184632, 186432, 189432, ...


5

The number $861432$ will work. I guess $123864$ looks nicer.


4

Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$. If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then its derivative is $2015ax^{2014}-2014bx^{...


3

$4\\ 24\\ 624\\ 3624\\ 183624$ 2,4,8 are good to work with, because once you find three digits at the end, you can put whatever you want in the front. 3,6,9 are good to work with too, because if you find something that divides by 3 or 9, you can shuffle the digits and it still divides by 3 (or 9). and 1 is just a no-brainer


3

Let us assume $(a^2 + b^2)/(ab + 1) = k \in \mathbb{N}$. We then have $a^2 - kab + b^2 = k$. Let us assume that $k$ is not an integer square, which implies $k \ge 2$. Now, we observe the minimal pair $(a, b)$ such that $a^2 - kab + b^2 = k$ holds. We may assume, without loss of generality, that $a \ge b$. For $a = b$, we get $k = (2 - k)a^2 \le 0$, so we ...


2

Your proof is almost valid, but it is unfortunately circular, because you use the following lemma: $$a \text{ is odd} \implies a^{n-1} \text{ is odd}$$ However, if you take the contrapositive of this statement: $$a^{n-1} \text{ is even} \implies a \text{ is even}$$ which is basically the same as your statement, except with $n-1$ instead of $n$. Therefore, ...


2

As Doug M said what you are doing is obtaining the modulo $9$ of the number, it is also called the Digital root of the number. An excerpt from Wikipedia: The digital root (also repeated digital sum) of a non-negative integer is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the ...


2

If $s$ is not a right zero divisor in $R'$, this is obviously impossible, since then $as=bs$ implies $a=b$ so $r'$ is the only element of $R'$ such that $r's=r$. Without this assumption, there are counterexamples. For instance, take $R=\mathbb{Z}$ and $R'=\mathbb{Z}[x]/(2x)$ and let $r=6$, $s=2$, and $r'=3+x$.


2

This is about 2-adic restrictions. First, odd squares of integers are $1 \pmod 8.$ Integer squares can only be $0,1,4 \pmod 8$ in any case. Therefore the sum of three integer squares cannot be $7 \pmod 8.$ Next, if the sum of three squares is divisible by $4,$ so $x^2 + y^2 + z^2 = k$ with $k \equiv 0 \pmod 4,$ then $x,y,z$ must be even so we can divide ...


1

Hint to get you started: If all of $a,b,c$ are multiples of $7$ then so is $n$ and we can divide all variables by $7$. Hence we may assume that not all of $a,b,c$ are multiples of $7$. A square can only be $\equiv 0,1,2,4\pmod 7$. For three of these to add to $0$, we must have $0+0+0$ (which we just excluded) or $1+2+4$.


1

Use the Euclidean algorithm on all $\frac{n(n-1)}{2}$ pairs of $a_i$ and $a_j$. If you get a GCD that is $F > 1$, then that is your answer. This is because $F \mid a_i$ and $F \mid a_j$, so when you multiply them together, you get $F^2 \mid (a_ia_j)$, so $F^2 \mid X$. Euclidean algorithm runs in $O(\log a_i+\log a_j)$, so since $a_i \leq 10^{18}$, it ...


1

Here's a software solution written in the c language using (obscure?) bit-wise logic checking if a number fulfills the criterion or not. int check_number(int a){ int l = a, mask = 0; for(;l;l/=10) { mask |= (1<<(l%10)); mask |= (!(mask%2)) && (a%(l%10) != 0);} return (__builtin_popcount(mask)==6) && (!(mask%2)); } The ...


1

Any divisor of $2^{\gcd(a,b)}-1$ is a divisor of $2^a-1$ since $a$ is a multiple of $\gcd(a,b)$ by definition. Hence $\;\gcd(2^a+1, 2^{\gcd(a,b)}-1)=\gcd(2^a+1, 2^a-1) =1$.


1

We $\gcd(a, b)$ is a divisor of $a$, so we have $a=m\gcd(a, b)$. Therefore, we can rephrase the question as: $$\gcd\left(2^{m\gcd(a, b)}+1, 2^{\gcd(a, b)}-1\right)$$ Now, since $1$ is a zero for $x^m-1$, so we know that: $$(x-1) \mid (x^m-1)$$ Substitute $x=2^{\gcd(a, b)}$: $$(2^{\gcd(a, b)}-1) \mid (2^{m\gcd(a, b)}-1)$$ Subtract the latter part of this ...


1

Let $ L = K(c_0, c_1, \ldots, c_n) $ be an extension of $ K $ by $ n+1 $ indeterminates, and consider the element $ g(x) = \sum c_k x^k $ in $ L[x] $. Let $ \bar{L} $ denote the splitting field of $ g $ over $ L $. It is clear that $ g $ has distinct roots in $ \bar{L} $, as otherwise no polynomial of degree $ n $ in $ K[x] $ could have distinct roots, and ...


1

Note that $(x+y)^j = x^j + y p_j(x,y)$ for some polynomial $p_j(x,y)$. With $y = f(x)$ you get $(x+y)^j = x^j + f(x) p_j(x,f(x))$. Now when you consider $f(x+f(x))$ use this. Collecting all the terms corresponding to the $x^j$ together yields just $f(x)$, while all the rest is clearly divisible by $f(x)$ as each summand was of the form $f(x)$ "times ...


1

See https://oeis.org/A091317 That deals with the case of $n$ prime (and in the text the case $n=p^2$. Thanks to @GerryMyerson for this (and the reference to AO14662). As he remarks, there is unlikely to be any useful characterisation known. So this is essentially an open problem. But the table does give you several more examples.


1

Given two integers $a$ and $b$, we say $a$ divides $b$ if there is an integer $c$ such that $b=ac$. Source. This is what $a$ divides $b$ means. The shorthand notation is $$a|b$$. In your example, $$a|a^2\iff a\leq a^2$$ since by definition there exists $c$ such that $a^2 = ac$, namely $a = c$.


1

$a$ is said to divide $b$ if there is an integer $c$ such that $bc=a$


1

much better to prove the contrapositive $a$ odd implies $a^n$ odd. This is easily proved by induction since the product of two odd numbers is odd.


1

It holds that for any number n the exponent $e_p(n)$ of a prime p in n! is exactly $e_p(n) = \frac{n-d_p(n)}{p-1}$ where $d_p(n)$ denotes the sum of digits in base p. This leads to the following reformulation of the problem: Find all positive integers $k$ such that for all positive integers n $\frac{(k-1)n-d_2(kn)+d_2(n)}{1}<(k-1)n+1 \\ d_2(n)-d_2(kn) ...


1

Hint $\ {\rm mod}\ 7\!:\ 5a\!-\!2\equiv 5(\color{#c00}{a\!+\!1})\,$ and $\,a^2\!-\!5a\!-\!6\equiv a^2\!+\!2a\!+\!1\equiv (\color{#c00}{a\!+\!1})^2$


1

Hint $\ \gcd(a,b)\, \mid\, a,b\, \mid\, {\rm lcm}\,(a,b)$ Thus $\ \gcd(a,b) \le a,b \le {\rm lcm}\,(a,b)$ So $ $ gcd = lcm $\, \Rightarrow \ a\!=\! b\ $ by squeezing.



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