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Question Number 222974    Answers: 1   Comments: 0

Question Number 222635    Answers: 0   Comments: 0

Question Number 222516    Answers: 2   Comments: 0

Given the integer k,how to find the incomplete general solution for the non-trivial integer solutions of the Diophantine equation: a^4 +b^4 +ka^2 b^2 =c^4 +d^4 +kc^2 d^2 ,a,b,c,d∈N,k∈Z,gcd(a,b,c,d)=1

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{integer}\:{k},\mathrm{how}\:\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{incomplete}\:\mathrm{general} \\ $$$$\mathrm{solution}\:\mathrm{for}\:\mathrm{the}\:\mathrm{non}-\mathrm{trivial}\:\mathrm{integer} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\:\mathrm{the}\:\mathrm{Diophantine}\:\mathrm{equation}: \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{ka}^{\mathrm{2}} {b}^{\mathrm{2}} ={c}^{\mathrm{4}} +{d}^{\mathrm{4}} +{kc}^{\mathrm{2}} {d}^{\mathrm{2}} ,{a},{b},{c},{d}\in\mathbb{N},{k}\in\mathbb{Z},\mathrm{gcd}\left({a},{b},{c},{d}\right)=\mathrm{1} \\ $$

Question Number 219732    Answers: 1   Comments: 0

Question Number 219657    Answers: 1   Comments: 0

prove; Π_(n=1) ^∞ (((5n−2)(5n−3))/((5n−1)(5n−4))) = ϕ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$

Question Number 218438    Answers: 1   Comments: 0

An amazing thing i saw S = 1 + 2 + 3 + 4 + 5 + 6... = 1 + 2(2/2 + 3/2 + 4/2 + 5/2 +6/2....) = 1 + 2(1 + 3/2 + 2 + 5/2 + 3...) = 1 + 2(1+ 2 + 3 ... + 3/2 + 5/2...) = 1 + 2S + 2Σ_(n= 1) ^∞ ((2n + 1)/2) or,S − 2S = 1 + Σ_(n=1) ^∞ 2n + 1 ∴ −S = Σ_(n=0) ^∞ 2n + 1 Sum of all odd numbers! I know the step S−2S = −S is not allowed

$${An}\:{amazing}\:{thing}\:{i}\:{saw} \\ $$$${S}\:=\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:\mathrm{5}\:+\:\mathrm{6}... \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{2}/\mathrm{2}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{4}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\mathrm{6}/\mathrm{2}....\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}\:+\:\mathrm{3}...\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}\left(\mathrm{1}+\:\mathrm{2}\:+\:\mathrm{3}\:...\:+\:\mathrm{3}/\mathrm{2}\:+\:\mathrm{5}/\mathrm{2}...\right) \\ $$$$\:\:\:\:=\:\mathrm{1}\:+\:\mathrm{2}{S}\:+\:\mathrm{2}\underset{{n}=\:\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$${or},{S}\:−\:\mathrm{2}{S}\:=\:\mathrm{1}\:+\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$$ \\ $$$$\therefore\:−{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}{n}\:+\:\mathrm{1} \\ $$$${Sum}\:{of}\:{all}\:{odd}\:{numbers}! \\ $$$${I}\:{know}\:{the}\:{step}\:{S}−\mathrm{2}{S}\:=\:−{S}\:{is}\:{not}\:{allowed} \\ $$

Question Number 218208    Answers: 1   Comments: 1

This question is really important Prove or disprove that lim_(n→∞) ((3^n m+3^(n−1) )/2^(⌈(n/2)⌉) ) + (3^(n−1) /2^n ) the limit exists for m ∈ N \B where B = {n ∣ log_2 (n) ∈ N }

$${This}\:{question}\:{is}\:{really}\:{important} \\ $$$${Prove}\:{or}\:{disprove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{n}} {m}+\mathrm{3}^{{n}−\mathrm{1}} }{\mathrm{2}^{\lceil\frac{{n}}{\mathrm{2}}\rceil} }\:+\:\frac{\mathrm{3}^{{n}−\mathrm{1}} }{\mathrm{2}^{{n}} }\: \\ $$$$\:{the}\:{limit}\:{exists}\:{for}\:{m}\:\in\:{N}\:\backslash{B} \\ $$$${where}\:{B}\:=\:\left\{{n}\:\mid\:{log}_{\mathrm{2}} \left({n}\right)\:\in\:{N}\:\right\} \\ $$

Question Number 218076    Answers: 2   Comments: 0

How many ways to arrnge the letters ABCCCDEFG (1) in general . (2) all 3 Cs must be together (3) only 2 Cs must be together (4) no 2 or 3 Cs be together (5) no letter still in its original place .

$${How}\:{many}\:{ways}\:{to}\:{arrnge} \\ $$$${the}\:{letters}\:{ABCCCDEFG} \\ $$$$\left(\mathrm{1}\right)\:{in}\:{general}\:. \\ $$$$\left(\mathrm{2}\right)\:{all}\:\mathrm{3}\:{Cs}\:{must}\:{be}\:{together} \\ $$$$\left(\mathrm{3}\right)\:{only}\:\mathrm{2}\:{Cs}\:{must}\:{be}\:{together} \\ $$$$\left(\mathrm{4}\right)\:{no}\:\mathrm{2}\:{or}\:\mathrm{3}\:{Cs}\:{be}\:{together} \\ $$$$\left(\mathrm{5}\right)\:{no}\:{letter}\:{still}\:\:{in}\:{its} \\ $$$${original}\:{place}\:. \\ $$

Question Number 217958    Answers: 2   Comments: 2

a,b,c∈Z^+ and a^2 +b^2 +c^2 +ab+bc+ca=2025 find out all triplets (a,b,c).

$${a},{b},{c}\in\mathbb{Z}^{+} {and} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ca}=\mathrm{2025} \\ $$$${find}\:{out}\:{all}\:{triplets}\:\left({a},{b},{c}\right). \\ $$

Question Number 217952    Answers: 2   Comments: 0

If x∈Z ∧y non-negative integer such that x^2 +10x+23=2^y find out x,y

$${If}\:{x}\in\mathbb{Z}\:\wedge{y}\:{non}-{negative}\:{integer} \\ $$$${such}\:{that} \\ $$$${x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{23}=\mathrm{2}^{{y}} \\ $$$${find}\:{out}\:{x},{y} \\ $$

Question Number 217244    Answers: 1   Comments: 0

Find all two-digit numbers such that when the number is divided by the sum of its digits the quotient is 4 and the remainder is 3.

$$\:\mathrm{Find}\:\mathrm{all}\:\mathrm{two}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{digits}\:\mathrm{the}\:\mathrm{quotient}\: \\ $$$$\mathrm{is}\:\mathrm{4}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{3}. \\ $$

Question Number 217132    Answers: 0   Comments: 0

Find all integers n> 1 such that n divides 2^(n−1) + 3^(n−1) .

$$ \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integers}\:\:\mathrm{n}>\:\mathrm{1}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{n}\:\:\mathrm{divides}\:\:\mathrm{2}^{\mathrm{n}−\mathrm{1}} \:+\:\mathrm{3}^{\mathrm{n}−\mathrm{1}} . \\ $$

Question Number 217130    Answers: 0   Comments: 0

Prove that for every integer n≥2 the number n^4 + 4^n is composite.

$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\:\mathrm{integer}\:\:\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{the}\:\mathrm{number}\:\:\mathrm{n}^{\mathrm{4}} +\:\mathrm{4}^{{n}} \:\:\mathrm{is} \\ $$$$\mathrm{c}{o}\mathrm{mposite}. \\ $$

Question Number 217129    Answers: 1   Comments: 2

prove that if an integer n is not divisible by 2 or 3 then n^2 ≡1(mod 24)

$${prove}\:{that}\:{if}\:{an}\:{integer}\:{n}\:{is}\:{not}\:{divisible}\:{by}\:\mathrm{2}\:{or}\:\mathrm{3} \\ $$$$\:{then}\:{n}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{24}\right) \\ $$

Question Number 217066    Answers: 1   Comments: 0

Find all integer x,y such that x^2 −y^2 =100

$${Find}\:{all}\:{integer}\:{x},{y}\:{such}\:{that} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{100} \\ $$

Question Number 217071    Answers: 1   Comments: 0

Find all two-digit numbers that are equal to four times the sum of their digits. Solve this using at least two different methods and verify your answers.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{two}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{that}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{four}\:\mathrm{times}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{their}\:\mathrm{digits}.\:\mathrm{Solve}\:\mathrm{this}\:\mathrm{using}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{different}\:\mathrm{methods}\: \\ $$$$\mathrm{and}\:\mathrm{verify}\:\mathrm{your}\:\mathrm{answers}. \\ $$

Question Number 217040    Answers: 2   Comments: 0

Find all positive integers n such that n divides 2^n + 1.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\mathrm{n}\:\:\mathrm{divides}\:\:\mathrm{2}^{{n}} \:+\:\mathrm{1}.\:\: \\ $$

Question Number 217030    Answers: 1   Comments: 0

Find all positive integers n such that n + 1 divides n^2 + 1

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\:\mathrm{n}\:\:\mathrm{such}\:\mathrm{that}\:\: \\ $$$$\:\mathrm{n}\:+\:\mathrm{1}\:\:\mathrm{divides}\:\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$

Question Number 216995    Answers: 3   Comments: 0

Find all prime numbers p and q such that p^2 − q^2 = 2024

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{numbers}\:\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\: \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{p}^{\mathrm{2}} −\:\:\mathrm{q}^{\mathrm{2}} =\:\:\mathrm{2024} \\ $$

Question Number 216912    Answers: 0   Comments: 0

Find all three-digit numbers n such that 1. n is divisible by the sum of its digits. 2. n is a perfect square.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{three}-\mathrm{digit}\:\mathrm{numbers}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{1}.\:{n}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\:\mathrm{of}\:\:\mathrm{its}\:\:\mathrm{digits}. \\ $$$$\mathrm{2}.\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Question Number 216911    Answers: 1   Comments: 0

Find all positive integer x,y such that x^2 + y^2 + xy = 169

$${Find}\:{all}\:{positive}\:{integer}\:\mathrm{x},\mathrm{y}\:{such}\:{that} \\ $$$$\mathrm{x}^{\mathrm{2}} +\:\mathrm{y}^{\mathrm{2}} +\:\mathrm{xy}\:=\:\mathrm{169} \\ $$

Question Number 216875    Answers: 1   Comments: 0

Let p be a prime number greater than 3. Prove that p^2 − 1 is always divisible by 24.

$$\mathrm{Let}\:\:\mathrm{p}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{p}^{\mathrm{2}} −\:\mathrm{1}\:\: \\ $$$$\mathrm{is}\:\:\mathrm{always}\:\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{24}. \\ $$

Question Number 216783    Answers: 3   Comments: 0

Find all positive integers n such that n^2 +7n+6 is perfect square.

$${Find}\:{all}\:{positive}\:{integers}\:{n}\:{such}\:{that} \\ $$$${n}^{\mathrm{2}} +\mathrm{7}{n}+\mathrm{6}\:{is}\:{perfect}\:{square}. \\ $$

Question Number 216769    Answers: 1   Comments: 0

Solve for integer k,m and n: k^2 m−n^2 =8

$${Solve}\:{for}\:{integer}\:{k},{m}\:{and}\:{n}: \\ $$$${k}^{\mathrm{2}} {m}−{n}^{\mathrm{2}} =\mathrm{8} \\ $$

Question Number 216664    Answers: 1   Comments: 0

Solve for non-negative integers: n^3 =3m(m+2n+1)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{non}-\mathrm{negative}\:\mathrm{integers}: \\ $$$$\:\:\:\mathrm{n}^{\mathrm{3}} =\mathrm{3m}\left(\mathrm{m}+\mathrm{2n}+\mathrm{1}\right) \\ $$

Question Number 216538    Answers: 0   Comments: 4

Find all integer solutions of 3^m =2n^2 +1. I only found m=1, 2, 5 by computer from m=1 to m=30000. Is there any greater solutions?

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{3}^{{m}} =\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}. \\ $$$$ \\ $$$${I}\:{only}\:{found}\:{m}=\mathrm{1},\:\mathrm{2},\:\mathrm{5}\:{by}\:{computer} \\ $$$${from}\:{m}=\mathrm{1}\:{to}\:{m}=\mathrm{30000}. \\ $$$${Is}\:{there}\:{any}\:{greater}\:{solutions}? \\ $$

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