Integer Definition and 606 Threads

One of my solution is: ##x^2-xy-6x+y^2+5y+6=(x-y)^2+(x-6)^2+(y+5)^2=49## and by trying integer ##(x,y)## points are ##(0,-2),(0,-3),(3,1),(3,-3),(4,-2),(4,1)## Another idea ##y^2+(5-x)y+x^2-6x+6=0## or ##x^2-(y+6)x+y^2+5y+6=0## and solving quadratic equation considering to integer solutions...

This is one of the things I've tried: import numpy as np import matplotlib.pyplot as plt import math num = int(input("Enter an integer: ")) math.factorial(num) print("The factorial of this value is " + num + ".") And this is the resulting error message I get: TypeError: can only...

This is not homework but just a mental exercise. I suspect there are no integer square roots in this set but can't prove it. I mean all digits are used once in each number. I am interested in arguments from principal and not algorithms. Note the number ##3## is missing. Thanks.

given something like: an = c where c is given and a, n, and c are only allowed to be integers. how would one find the value of say n or a?

I am really struggling in how to begin this problem. So far I have considered using the Euclidean Algorithm and trying to find the gcd of each number like gcd(9,10) but each time they give me 1 so that doesn't work. My next idea is to do a proof by contradiction where I start with assuming that...

Hello, Am re-studying math & calculus aiming to start pure math studying later. However, I got this problem in Stewart calculus. Typically, this is a straightforward IVT application. x = x^3 + 1, call f(x)= x^3 - x + 1 & apply IVT. However I have two things to discuss. First thing is simple...

Let ## 5k+4=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}} ##. Then ## 5\equiv 0\pmod {5} ## and ## 5k+4\equiv 4\pmod {5} ##. Thus ## p_{i}^{k_{i}}\not \equiv 0\pmod {5} ## for ## i=1, 2,..., s ##. Suppose all ## p_{i}^{k_{i}}\equiv 1\pmod {5} ##. Then ## p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb...

Let ## x ## be an integer. Then ## x\equiv 1\pmod {2}, x\equiv 2\pmod {3}, x\equiv 5\pmod {6} ## and ## x\equiv 5\pmod {12} ##. Note that ## x\equiv 5\pmod {6}\implies x\equiv 5\pmod {2\cdot 3} ## and ## x\equiv 5\pmod {12}\implies x\equiv 5\pmod {3\cdot 4} ##. Since ## gcd(2, 3)=1 ## and ##...

Let ## x ## be an integer. Then ## x\equiv 2\pmod {3}, x\equiv 3\pmod {4}, x\equiv 4\pmod {5} ## and ## x\equiv 5\pmod {6} ##. This means \begin{align*} &x\equiv 2\pmod {3}\implies x+1\equiv 3\pmod {3}\implies x+1\equiv 0\pmod {3},\\ &x\equiv 3\pmod {4}\implies x+1\equiv 4\pmod {4}\implies...

Consider a certain integer between ## 1 ## and ## 1200 ##. Then ## x\equiv 1\pmod {9}, x\equiv 10\pmod {11} ## and ## x\equiv 0\pmod {13} ##. Applying the Chinese Remainder Theorem produces: ## n=9\cdot 11\cdot 13=1287 ##. This means ## N_{1}=\frac{1287}{9}=143, N_{2}=\frac{1287}{11}=117 ## and...

Let ## a>2 ## be the smallest integer. Then \begin{align*} &2\mid a\implies a\equiv 0\pmod {2}\implies a\equiv 2\pmod {2}\\ &3\mid (a+1)\implies a+1\equiv 0\pmod {3}\implies a\equiv -1\pmod {3}\implies a\equiv 2\pmod {3}\\ &4\mid (a+2)\implies a+2\equiv 0\pmod {4}\implies a\equiv -2\pmod...

Consider the integer ## 1010908899 ##. Observe that ## 7\cdot 11\cdot 13=1001 ##. Then ## 10^{3}\equiv -1\pmod {1001} ##. Thus \begin{align*} &1010908899\equiv (1\cdot 10^{9}+10\cdot 10^{6}+908\cdot 10^{3}+899)\pmod {1001}\\ &\equiv (-1+10-908+899)\pmod {1001}\\ &\equiv 0\pmod {1001}.\\...

Proof: Suppose that ## 6 ## divides ## N ##. Let ## N=a_{m}10^{m}+\dotsb +a_{2}10^{2}+a_{1}10+a_{0} ##, where ## 0\leq a_{k}\leq 9 ##, be the decimal expansion of a positive integer ## N ##. Note that ## 6=2\dotsb 3 ##. This means ## 2\mid 6 ## and ## 3\mid 6 ##. Then ## 2\mid...

Proof: Suppose ## 2^{n} ## divides an integer ## N ##. Let ## N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0} ## for ## 0\leq a_{k}\leq 9 ##. Then ## 2^{n}\mid N\implies 2^{n}\mid (a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0}) ##. Note that ## 10^{k}=2^{k} 5^{k}\equiv 0\pmod {2^{n}} ##...

Proof: Let ## a ## be any integer. Then ## a\equiv 0, 1, 2, 3, 4, 5, 6, 7 ##, or ## 8\pmod {9} ##. This means ## a^{2}\equiv 0, 1, 4, 9, 7, 7, 0, 4 ##, or ## 1\pmod {9} ## and ## a^{3}\equiv 0, 1, 8, 0, 1, 8, 0, 1 ##, or ## 8\pmod {9} ##. Thus ## a^{2}\equiv 0, 1, 4 ##, or ## 7\pmod {9} ## and...

Proof: Let ## P(x)= \Sigma^{m}_{k=0} a_{k} x^{k} ## be a polynomial function. Then ## N=a_{m}10^{m}+a_{m-1}10^{m-1}+\dotsb +a_{1}10+a_{0} ## for ## 0\leq a_{k}\leq 9 ##. Since ## 10\equiv 1\pmod {3} ##, it follows that ## P(10)\equiv P(1)\pmod {3} ##. Note that ## N\equiv (a_{m}+a_{m-1}+\dotsb...

Proof: Suppose ## N ## is the integer and ## x ## is the units digit of ## N ##. Then ## N=10k+x ## for some ## k\in\mathbb{Z} ## where ## x={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ##. Note that ## 10k\equiv 0\pmod {2}\implies N\equiv x\pmod {2} ##. Thus ## 2\mid N\implies N\equiv 0\pmod {2}\implies...

Observe that ## (447836)_{9}=6+3\cdot 9+8\cdot 9^{2}+7\cdot 9^{3}+4\cdot 9^{4}+4\cdot 9^{5}=268224 ##. Then ## 2+6+8+2+2+4=24 ##. Thus ## 3\mid (2+6+8+2+2+4)\implies 3\mid (447836)_{9} ## and ## 8\mid (2+6+8+2+2+4)\implies 8\mid (447836)_{9} ##. Therefore, the integer ## (447836)_{9} ## is...

Proof: Let ## a ## be any integer. Then ## a\equiv 0, 1, 2, 3, 4, 5, 6, 7, 8 ##, or ## 9\pmod {10} ##. Note that ## a^{2}\equiv 0, 1, 4, 9, 6, 5, 6, 9, 4 ##, or ## 1\pmod {10} ##. Thus ## a^{2}\equiv 0, 1, 4, 5, 6 ##, or ## 9\pmod {10} ##. Therefore, the units digit of ## a^{2} ## is ## 0, 1...

Proof: Suppose ## a\equiv b\pmod {n_{1}} ## and ## a\equiv c\pmod {n_{2}} ## where the integer ## n=gcd(n_{1}, n_{2}) ##. Then ## a-b=n_{1}k_{1} ## and ## a-c=n_{2}k_{2} ## for some ## k_{1}, k_{2}\in\mathbb{Z} ##. This means ## b-c=n_{2}k_{2}-n_{1}k_{1} ##. Since ## n=gcd(n_{1}, n_{2}) ##, it...

## y-x \gt 1 \implies y \gt 1+x## Consider the set ##S## which is bounded by an integer ##m##, ## S= \{x+n : n\in N and x+n \lt m\}##. Let's say ##Max {S} = x+n_0##, then we have $$ x+n_0 \leq m \leq x+(n_0 +1)$$ We have, $$ x +n_0 \leq m \leq (x+1) +n_0 \lt y+ n_0 $$ Thus, ##x+n_0 \leq m \lt...

Proof: Suppose that the integer ## a ## is not divisible by ## 2 ## or ## 3 ##. Then ## a\equiv 1, 5, 7, 11, 13, 17, 19 ## or ## 23\pmod {24} ##. Note that ## a\equiv b\pmod {n}\implies a^{2}\equiv b^{2}\pmod {n} ##. Thus ## a^{2}\equiv 1, 25, 49, 121, 169, 289, 361 ## or ## 529\pmod...

Proof: Suppose ## a ## is an odd integer. Then ## a=2k+1 ## for some ## k\in\mathbb{Z} ##. Note that ## a^{2}=(2k+1)^{2}=4k^{2}+4k+1=4k(k+1)+1 ##. Since ## k(k+1) ## is the product of two consecutive integers, it follows that ## k(k+1) ## must be even. This means ## k(k+1)=2m ## for some ##...

The last three digits of ##x^3## must be solely dependent on the last 3 digits of ##x##. So let ##x=a+10b+100c## for integers ##a,b,c##. Then ##x^3 = a^3 + 30 a^2 b + 300 a b^2 + 300 a^2 c +O(1000)## where of course ##O(1000)## don't affect the last 3 digits. Evidently ##a^3## is the only...

Proof: First, we will prove that the integer ## 53^{103}+103^{53} ## is divisible by ## 39 ##. Note that ## 53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39} ##. Now observe that ## 103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv...

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{2021}$$ $$\frac{y+x+1}{xy}=\frac{1}{2021}$$ $$xy = 2021y + 2021 x + 2021$$ Then I am stuck. How to continue? Thanks

Distance between point (-4, 5) and point on circle: $$d=\sqrt{(x+4)^2+(y-5)^2}$$ $$=\sqrt{x^2+8x+16+y^2-10y+25}$$ Then substitute ##y^2## from equation of circle: $$d=\sqrt{x^2+8x+16-x^2+4x-6y+12-10y+25}$$ $$=\sqrt{12x-16y+53}$$ After this, I need to try the points one by one to check whether...

Closer to odd number implies ##|y/x - (2n+1)| < 1/2## for ##n = 0,1,2...##. Then $$-\frac 1 2 < \frac y x - (2n+1) < \frac 1 2 \implies\\ y < (2n + 1.5)x,\\ y > (2n + 0.5)x$$ for each ##n##. We note ##x \in (0,1)## implies ##y## can be larger than 1 since the slope is greater than 1 (but we know...

Proof: Let ## a>5 ## be an integer. Now we consider two cases. Case #1: Suppose ## a ## is even. Then ## a=2n ## for ## n\geq 3 ##. Note that ## a-2=2n-2=2(n-1) ##, so ## a-2 ## is even. Applying Goldbach's conjecture produces: ## 2n-2=p_{1}+p_{2} ## as a sum of two primes ## p_{1} ## and ##...

Proof: Let ## n ## be an integer. Then ## 2n=p_{1}+p_{2} ## for ## n\geq 2 ## where ## p_{1} ## and ## p_{2} ## are primes. Suppose ## n=k-1 ## for ## k\geq 3 ##. Then ## 2(k-1)=p_{1}+p_{2} ## ## 2k-2=p_{1}+p_{2} ## ## 2k=p_{1}+p_{2}+2 ##. Thus ## 2k+1=p_{1}+p_{2}+3 ##...

Proof: Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##, which is an odd integer that is not greater than ## n ##. Let ## n>1 ## be an integer. Since ## n! ## is even, it follows that ## n!+1 ## is odd. Thus ## 2\nmid (n!+1) ##. This means every prime factor...

Proof: Suppose ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational. Then we have ## \sqrt[n]{a}=\frac{b}{c} ## for some ## b,c\in\mathbb{Z} ## such that ## gcd(b, c)=1 ## where ## c\neq 0 ##. Thus ## \sqrt[n]{a}=\frac{b}{c} ## ## (\sqrt[n]{a})^{n}=(\frac{b}{c})^n ##...

Proof: Suppose ## n>1 ## is a positive integer. Let ## n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}} ## be the prime factorization of ## n ## such that each ## k_{i} ## is a positive integer and ## p_{i}'s ## are prime for ## i=1,2,3,...,r ## with ## p_{1}<p_{2}<p_{3}<\dotsb <p_{r} ##...

Proof: Consider all primes ## p\leq\sqrt{701}\leq 27 ##. Note that ## 701=2(350)+1 ## ## =3(233)+2 ## ## =5(140)+1 ## ## =7(100)+1 ## ## =11(63)+8 ## ## =13(53)+12 ##...

Proof: Suppose an integer ## n>1 ## is square-free. Then we have ## a^2\nmid n, \forall a\in\mathbb{Z} ##. Let ## n=p_{1}^{a_{1}} p_{2}^{a_{2}}\dotsb p_{r}^{a_{r}} ## be the prime factorization of ## n ## such that each ## a_{i} ## is a positive integer and ## p_{i}'s ## are prime for ##...

Proof: Suppose a positive integer ## a>1 ## is a square. Then we have ## a=b^2 ## for some ## b\in\mathbb{Z} ##, where ## b=p_{1}^{n_{1}} p_{2}^{n_{2}} \dotsb p_{r}^{n_{r}} ## such that each ## n_{i} ## is a positive integer and ## p_{i}'s ## are prime for ## i=1,2,3,...,r ## with ##...

Proof: Suppose ## n>1 ## is an integer not of the form ## 6k+3 ##. Then we have ## n=6k ## for some ## k\in\mathbb{Z} ##. Thus ## n^{2}+2^{n}=(6k)^{2}+2^{6k} ## ## =36k^{2}+2^{6k} ## ## =2(18k^{2}+2^{6k-1}) ##...

Proof: Suppose for the sake of contradiction that ## p=2^{k}-1 ## is prime but ## k ## is not an odd integer. That is, ## k ## is an even integer. Then we have ## k=2a ## for some ## a\in\mathbb{Z} ##. Thus ## p=2^{k}-1 =2^{2a}-1 =4^{a}-1. ## Note that ## 3\mid...

Proof: Suppose n is an integer such that ## n>11 ##. Then n is either even or odd. Now we consider these two cases separately. Case #1: Let n be an even integer. Then we have ## n=2k ## for some ## k\in\mathbb{Z} ##. Consider the integer ## n-6 ##. Note...

Proof: Suppose ##a=8^n+1 ## for some ##a \in\mathbb{Z}## such that n##\geq##1. Then we have ##a=8^n+1 ## =## (2^3)^n+1 ## =## (2^n+1)(2^{2n} -2^n+1) ##. This means ## 2^n+1\mid 2^{3n} +1 ##. Since ##2^n+1>1## and ##2^{2n} -2^n+1>1## for all...

Proof: Suppose a=n^4+4 for some a##\in\mathbb{Z}## such that n>1. Then we have a=n^4+4=(n^2-2n+2)(n^2+2n+2). Note that n^2-2n+2>1 and n^2+2n+2>1 for n>1. Therefore, every integer of the form n^4+4, with n>1, is composite.

Proof: First, we will show that gcd(a, 0)=abs(a). Suppose a is a nonzero integer such that a##\neq##0. Note that gcd(a, 0)##\le##abs(a) by definition of the greatest common divisor. Since abs(a) divides both a and 0, we have that...

Is it always true that the square root of an odd powered integer will always be irrational?

Find the smallest positive integer N that satisfies all of the following conditions: • N is a square. • N is a cube. • N is an odd number. • N is divisible by twelve prime numbers. How many digits does this number N have?Please Explain your steps in detail.

Look at this example, >>> a = 97 >>> type(a) <class 'int'> >>> bin(a) '0b1100001' >>> b = ord('a') >>> b 97 >>> type(b) <class 'int'> >>> bin(b) '0b1100001' Is this means that the string 'a' and the integer 97 stored as the same binary in the memory ? If so then how can python tell the...

Compute the number of positive integer divisors of 10!. By the fundamental theorem of arithmetic and the factorial expansion: 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 2 x 5 x 3^2 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2 x 1 = 2^8 x 3^4 x 5^2 x 7 Then there are 9 possibilities for 2, 5 for 3, 3 for...

i do not seem to understand part ##ii## of this problem...mathematical induction proofs is one area in maths that has always boggled me :oldlaugh: let ##n=3, p=7, ⇒m=4## therefore, ##7=(4-3)(4+3)## ##7=1⋅7## ##1, 7## are integers...##p## is prime. i am attempting part ##iii## in a moment...

Hello, I was wondering how to prove that the Binomial Series is not infinite when k is a non-negative integer. I really don't understand how we can prove this. Do you have any examples that can show that there is a finite number when k is a non-negative integer? Thank you!

Find all positive integers $n$ for which the equation $(x^2+y^2)^n=(xy)^{2014}$ has positive integer solutions.

Let $x$ be a real number such that $x=(x-1)^3$. Show that there exists an integer $N$ such that $-2^{1000}<x^{2021}-N<2^{-1000}$.