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Hint: Apply the rational root theorem to $$x^2 - ( a + b) x + ab$$

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If $d|a$, then $a = dn$ for some integer $n$. If $d|b$, then $b = dm$ for some integer $m$. Multiply $a$ and $b$ together: $ab = (dn)(dm) = d^2mn$ which is exactly what $d^2|ab$ means.

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Modest progress. There are infinitely many integers $n$ such that $n^3+1\mid n!$. We always have $n^3+1=(n+1)(n^2-n+1)$. Let $n=k^2+1$. Then $$n^2-n+1=(1+k+k^2)(1-k+k^2).$$ Assume further that $k\equiv1\pmod3$. In that case $1+k+k^2$ and $n+1=2+k^2$ are both divisible by $3$. For all sufficiently large $k\equiv1\pmod3$ we thus have $$... 11 Using modular arithmetic: If x \equiv 0 \mod 4, then x^3 \equiv 0 \mod 4 and x^3-2 \equiv 2 \mod 4. If x \equiv 1 \mod 4, then x^3 \equiv 1 \mod 4 and x^3-2 \equiv 3 \mod 4. If x \equiv 2 \mod 4, then x^3 \equiv 8 \equiv 0 \mod 4 and x^3-2 \equiv 2 \mod 4. If x \equiv 3 \mod 4, then x^3 \equiv 27 \equiv 3 \mod 4 and x^3-2 \equiv ... 8 It suffices to show that for infinitely many n, the largest prime factor of n^{2015}+1 is at most \sqrt{n}. Indeed, if n is such a large integer and p is a prime, then the largest value of a for which p^a\mid n^{2015}+1 is \leq c \log n for some constant c, while n! is divisible by p^a with a\geq \frac{n}{p}-1\geq \sqrt{n}-1>c \log ... 7 HINT: If n\equiv0,\pm1,\pm2,\pm3;n^2\equiv0,1,4,2\pmod7 So, a^2,b^2\equiv1,2,4 Check for c^2\pmod7 when a^2\not\equiv b^2\pmod7 But my greater concern is how the problem, specifically \pmod7 was conceived! 7 The answer is yes, because 4^2\equiv 4\pmod{6}, and hence 4^n\equiv 4\pmod{6} for all n\ge 1. 6 n^3+3n^2+2n=(n+1)(n+2)n. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both 2 and 3 and hence is divisible by the least common multiple of 2 and 3, which turns out to be 6. 5 Using Euclid's formula, a=2mn, b=m^2-n^2 We have 7\nmid2mn(m^2-n^2) Now, (m^2-n^2)^2-(2mn)^2=m^4+n^4-6m^2n^2\equiv m^4+n^4+m^2n^2\pmod7 But (m^2-n^2)(m^4+n^4+m^2n^2)=(m^2)^3-(n^2)^3\equiv1-1\pmod7 using Fermat's Little Theorem as (m,7)=(n,7)=1 \implies7|(m^4+n^4+m^2n^2) as 7\nmid(m^2-n^2) Can you take it from here? 5 In problem (a), use Fermat's little theorem, which says (or a rather, a very slightly different version says) that for any prime number p, and any integer n that's not divisible by p, we have$$n^{p-1}\equiv 1\bmod p$$In particular, use n=2 and p=17. Keep in mind that 2017=(126\times 16)+1. In problem (b), note that 30\equiv 61\equiv -1\bmod ... 5 Hints: a) 2016 is divisible by 16. Now use Fermat's Theorem. b) 30\equiv -1\pmod{31} and 61\equiv -1\pmod{31}. c) If the prime p is not equal to 3 then p^2\equiv 1\pmod{3}. 5 Let’s look at the non-negative integers having base ten representations requiring at most d digits. (Including 0 makes no different in the limit and makes the calculation simpler.) There are 10^d of them. Of those, 9^d have representations without a 7. Thus, the fraction of these numbers with a 7 in their representations is ... 4 Let p^4-20p^2+19 = (p^2-1)(p^2-19), and 180 = 6^2\cdot 5. But p\equiv \pm 1\pmod{6}, so p^2\equiv 1\pmod{6}, thus p^2-1 \equiv 1-1 = 0 \pmod{6} and p^2-19 \equiv 1-19 = -18 \equiv 0 \pmod{6}, therefore 6^2 \mid p^4-20p^2+19. On the other hand p\not \equiv 0\pmod{5}, so p^2\equiv \pm 1\pmod{5} and p^4\equiv 1\pmod{5}, thus p^4-20p^2 + 19 ... 4 Since 2015 = 5\cdot 13\cdot 31 , and n^a + 1| n^{ab}+1  if b is odd, a necessary condition for n^{2015}+1 | n!  is n^m+1 | n! for every m in \{5, 13, 31 , 5\cdot 13 , 5\cdot 31 , 13\cdot 31 \} . Solutions are going to be hard to find. All those expressions of the form n^j-n^{j-1}+...-n+1  for odd j will have to have all prime factors \le ... 4 Let n:=a+b and m:=ab. Then a^2-an+m=0 and hence$$a=\dfrac{n\pm\sqrt{n^2-4m}}{2}.$$note that since a\in\Bbb Q, then n^2-4m must be a perfect square. Then necessarily 2\mid n\pm\sqrt{n^2-4m} (if n is odd, then n^2-4m is odd \Longrightarrow n\pm\sqrt{n^2-4m} is even; if n is even, then n^2-4m is even \Longrightarrow ... 3 For all n,$$ 4^n\equiv1^n\equiv1\pmod{3} $$For all n\gt0,$$ 4^n\equiv0^n\equiv0\pmod{2} $$Therefore, by the Chinese Remainder Theorem$$ 4^n\equiv4\pmod{6} $$and therefore, for all n\gt0,$$ 4^n+2\equiv0\pmod{6} $$3$$x^3\equiv2\pmod4\implies x^3\equiv2\pmod2\equiv0as x^3-x=x(x-1)(x+1)\equiv0\pmod2, x^3\equiv x\pmod2 \implies x is even \implies x^3\equiv0\pmod4 3 If k is odd x^3 is odd and x^3-2 also, so 4 can't divide it. If x is even it is 2k and so x^3-2=(2k)^3-2=8k^3-2=4(2k^3)-2 and so it is not a multiple of 4 3 A) \begin{align*} &\,(2a+1)^2+(2b+1)^2=(4a^2+4a+1)+(4b^2+4b+1)=4(a^2+b^2)+4(a+b)+2\\ =&\,2[2(a^2+b^2)+2(a+b)+1]. \end{align*} Since 2(a^2+b^2) and 2(a+b) are both even, the expression between the brackets is odd because of the +1 term. Now, the double of an odd number can never be a perfect square. (Try proving this last statement.) B) ... 3 Hint For part A), it may be helpful to first prove the following: n^2 \equiv 0~\text{or}~1\mod{4} for all integers n. 3 Let d = \gcd(a, b), and let a = dA and b = dB. Then, since \operatorname{lcm}(a, b) =\frac{ab}{\gcd(a, b)} =\frac{ABd^2}{d} =ABd , d+ABd =Ad+Bd , so 1+AB =A+B  or 0 =AB-A-B+1 =(A-1)(B-1) . Therefore either A=1 or B=1. If A=1, then a = dA = d divides b = dB. Similarly, if B = 1, b divides a. 3 180=5\cdot9\cdot4 For any integer p, p^4-20p^2+19\equiv p^4-2p^2+1\pmod9 Now p^4-2p^2+1=(p^2-1)^2 For (p,3)=1,p\equiv\pm1\pmod3\implies p^2\equiv1\pmod3\iff p^2-1\equiv0 and for any integer p, p^4-20p^2+19\equiv p^4-1\pmod{20} \implies p^4\equiv1\pmod5 by Fermat's Little Theorem for (5,p)=1 Now if (p,2)=1 p\pm 1 are even,\implies ... 3 Just write every term in the sum in terms of a_1 and a_2 (keeping in mind that a_{n+2}=a_n+a_{n+1}):a_1+a_2+(a_1+a_2)+(a_1 + 2a_2)+(2a_1+3a_2)+(3a_1+5a_2)+(5a_1+8a_2)+(8a_1+13a_2)+(13a_1+21a_2)+(21a_1+34a_2). $$Then the sum is clearly equal to 55a_1+88a_2 = 11(5a_1+8a_2), which is 11 times the seventh term of the sum. 2 Take a common divisor k of a and b. That means that we can write a=ka', b=kb'. Then, if d=ax+by,$$d=a'kx+b'ky=k(a'x+b'y)$$so we see that k also divides d. Since \gcd(a,b) is a common divisor of a and b (namely, the greatest), it also holds the property. Actually, the idea behind this is very simple: the sum of two multiples of the ... 2 I dont understand programming, but what I can tell you is \frac{a+b}{c+d} \neq \frac{\frac{a}{b}+\frac{c}{d}}{2}, that is why it didn't work. What you can do is may be, try \frac{a+b}{c+d}=\frac{\frac{a}{b}+1}{\frac{c}{d}+1}.\frac{b}{d}. In you example a=12,b=6,c=5,d=2 which will give you 2.57 2$$(a^2+3)(a^2+7)=(a^2-1+4)(a^2-1+8)=\{(a-1)(a+1)\}^2+12(a-1)(a+1)+32$$Now as a is odd, one of a-1,a+1 is divisible by 4 and the other is by 2, not by 4 \implies 4\cdot2 divides (a-1)(a+1) 2 In fact, it does not work for all k, x \in \mathbb{Z}. Consider k = 4, x = 2: 4 does not divide 2^4 - 2 = 16 - 2 = 14. However, this works if k is prime, and is well-known Fermat's little theorem. 2 In \mathbf Z/4\mathbf Z, the squares are 0 and 1, hence the cubes are 0,1,-1, not 2. 2 First remember the obvious fact that 4 = 2 \times 2, then everything else falls into place. If x = 2y (meaning that x is even), then x^3 = (2y)^3 = 8y^3. Then$$\frac{x^3 - 2}{4} = \frac{8y^3 - 2}{4} = 2y^3 - \frac{1}{2}, which is clearly not an integer. If $x$ is odd, I'm not telling you anything new with that $x^3$ is odd also, as well as $x^3 ... 2 As you point out, a 7-digit number cannot have a 5 in it (otherwise, it would have to end in 5, but would have to have one of 2, 4, 6, or 8 in it, so must be even). Thus a 7-digit number$x\$ contains 7 of 1, 2, 3, 4, 6, 7, 8, 9. We need only identify the missing digit. If it is not 9, the resulting number must be divisible by 9, so the missing digit must be ...

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