(we want to find a p*q=a²+2ab+b; with p and q ∈ N*) So, we will star isolating b and a in the equation to find a divisible term after: For every b and a, not necessarly in Z: a²+2ab+b=44 a²+b(2a+1)=44 2a²+a+b(2a+1)=44+a²+a a(2a+1)+b(2a+1)=44+a²+a (a+b)(2a+1)=44+a²+a a+b= [a²+a+44]/(2a+1) ● b = {[a²+a+44]/(2a+1)}-a; with 2a+1≠0 But, remember b ∈ N*, then: make it: Q(a)=a²+a+44 and q(a)=2a+1 then, [a²+a+44]/(2a+1)=[Q(a)]/[q(a)]=p(a) and p(a)=n, n ∈ N*. /////////////////////////////////////////////////// finding the p(a) and r(a): Q(a) = a² + a + 44 | q(a) =(2a+1) -a² - a/2 ||p(a)=(a/2)+(1/4) -----------‐ 0 + a/2 + 44 - a/2 - 1/4 ------------------- 0 + 175/4 = r(a) ///////////////////////////////////////////////// Constructing p*q into the original equation: So, Q(a)=q(a)*p(a) + r(a) ● a²+a+44=(2a+1)*[(a/2)+(1/4)] +(175/4) retake the original equation: a²+2ab+b=44 a²+b(2a+1)=44 2a²+a+b(2a+1)=44+a²+a a(2a+1)+b(2a+1)=44+a²+a (a+b)(2a+1)=44+a²+a; but remember Q(a)=44+a²+a then, (a+b)(2a+1)=(2a+1)[(a/2)+(1/4)]+(175/4) make 4 times the equation: (4a+4b)(2a+1)=(2a+1)[2a+1]+175 (4a+4b)(2a+1)=(2a+1)²+175 (4a+b)(2a+1) - (2a+1)² =175 (2a+1)[(4a+4b)-(2a+1)]=175 (2a+1)[4a-2a+4b-1]=175 (2a+1)(2a+4b+1)=175 Where are the p and q? It is here the term p=(2a+1) and it the term q=[(2a+1)+4b]. ■(2a+1)[(2a+1)+4b]=175; n=2a+1 better make like this way: ■n(n+4b)=175 Now, we can thing the possibilites: n×(n+4b)=5²×7; with ■D(175)={1;5;7;25;35;175} And ■ n ⊂ D(175) (I) then, n ⊂ {1;5;7;25;35;175} ////////////////////////////////////////////////// Considering the domain of the question: remember that n=2a+1 and consider a ∈ N* because the question already said it. And, remember b ∈ N. And think about a domain of a and b. So, retake a²+2ab+b=44 you can see that each term of the equation cannot be bigger than 44. Then, a²
I like your factorization solution, but I think this can be solved simply in this way(not sure is acceptable for math olympiads): since a>0,b>0 than a^2b=8,a=3=>b=5
I would go in a similar way. Factorizing is a pain in the ass so I would just separate b b=(-a^2+44)/(2a+1) |X2 2b=(-a(2a+1-1)+88)/(2a+1) |X2 4b+2a=(2a+1-1+176)/(2a+1) 4b+2a-1=175/(2a+1) (4b+2a-1)(2a+1)=175 and from this we can look for all possible positive or negative pairs which can satisfy the equation
First,dont use a line for division when you are multiplying!!This confuses students.In addition ,this i s much easier solved by trial and error.We know a is limited between 1 and 6.Just try them!!
@SALogics DEGERLİ dostum tüm üslü sayılar mod la çözülebilir.Bunun çalışmasını yapıyorum .Aslında asal sayılarla ilgilenirken üslü sayılar tesadüfen karşıma çıktı ve logaritma limit trigonometri 10 yaşındaki çocugunn çözebileceği yöntemler buldum.😁😁Örneğin x^3+y^3+z^3=a^3 mü sorusu cevabı irrasyonel olarak vardır.
If the original equation is transformed into an equation that has the one variable on one side of the equation and the other variable on the opposite side of the equation, we will have: b=(44-a^2)/(2a+1). From the above relationship, if we assign any value for "a", we will obtain a corresponding value for "b". Therefore, there are infinitely many pair of solutions for "a" and "b". Examples are: (a,b)=(0,44), (2,8), (3,5), (12,-4), (17,-7), (87,-43), (-1,-43),(-3,-7), (-4,-4), (-13,5), (-18,8), (-88,44) and so on.
@@ronbannon Thanks. I agree with you that there are only a finite number of integer solutions. However, the original problem is presented in a way that it simply asks for a=?, b=?. No restrictions are mentioned beforehand that a, b must be integer values. It is only when you click to play the video will you see the restrictions (a>0 and b>0). As a huge fan of these mathematical puzzles, I always try to solve the problems presented using my own analysis and approach. After I've done that, that's the time I play the video to verify if my answers are correct. Just to refresh you, notice that the original equation presented is actually an equation of a parabola (where the variables "a" and "b" are used instead of "x" and "y" found in analytic geometry textbooks. Since the parabola is a continuous locus of points, there would be an infinite number of points along its curve (unless you impose a restriction on the range and nature of values for "x" and "y".
@@ronbannon Thanks. I agree with you that there are only a finite number of integer solutions. However, the original problem is presented in a way that it simply asks for a=?, b=?. No restrictions are mentioned beforehand that a, b must be integer values. It is only when you click to play the video will you see the restrictions (a>0 and b>0). As a huge fan of these mathematical puzzles, I always try to solve the problems presented using my own analysis and approach. After I've done that, that's the time I play the video to verify if my answers are correct. Just to refresh you, notice that the original equation presented is actually an equation of a parabola (where the variables "a" and "b" are used instead of "x" and "y" found in analytic geometry textbooks. Since the parabola is a continuous locus of points, there would be an infinite number of points along its curve (unless you impose a restriction on the range and nature of values for "x" and "y".
@@SALogics😳🤣😡 a,b є N => a=1,2,3... a,b є Z+ => a=0,1,2,3... 😬 The day before: @habeebalbarghothy6320 Also (a,b)=(0,44) is possible. @SALogics. ❤️ Yes, you are right!❤️ 😂 Verify: 0²+2•0•44+44=44 !!!
The main reason u are not getting support even after such good content is that your channel name is salogic, but if someone searches sa logic, then your channel does not appear, so plz get a better name for your channel related to maths..For example Maths GO!
rearrange the eq to: b = [44-a^2] / [2a + 1]. since both a, b > 0 a
Very nice! ❤
a²+2ab+b²=44+b²-b
Only b=5 and b=8 give perfect square. Then calculate a.
Very nice! ❤
(2a+1)b=44-a^2>0.a=1,2,3,4,5,6. (a,b)=(2,8),(3,5)
Very nice! ❤
(2a+1)b=44-a^2 . 2a+1>=3
b=(44-a^2)/(2a+1)=(1/4)*{(-2a+1)+175/(2a+1)}
(2a+1)(2a+4b-1)=175
(2a+1,2a+4b-1)=(5, 35),(7, 25)
Very nice! ❤
(we want to find a p*q=a²+2ab+b; with p and q ∈ N*)
So, we will star isolating b and a in the equation to find a divisible term after:
For every b and a, not necessarly in Z:
a²+2ab+b=44
a²+b(2a+1)=44
2a²+a+b(2a+1)=44+a²+a
a(2a+1)+b(2a+1)=44+a²+a
(a+b)(2a+1)=44+a²+a
a+b= [a²+a+44]/(2a+1)
● b = {[a²+a+44]/(2a+1)}-a;
with 2a+1≠0
But,
remember b ∈ N*, then:
make it:
Q(a)=a²+a+44 and q(a)=2a+1
then,
[a²+a+44]/(2a+1)=[Q(a)]/[q(a)]=p(a)
and p(a)=n, n ∈ N*.
///////////////////////////////////////////////////
finding the p(a) and r(a):
Q(a) = a² + a + 44 | q(a) =(2a+1)
-a² - a/2 ||p(a)=(a/2)+(1/4)
-----------‐
0 + a/2 + 44
- a/2 - 1/4
-------------------
0 + 175/4 = r(a)
/////////////////////////////////////////////////
Constructing p*q into the original equation:
So,
Q(a)=q(a)*p(a) + r(a)
● a²+a+44=(2a+1)*[(a/2)+(1/4)] +(175/4)
retake the original equation:
a²+2ab+b=44
a²+b(2a+1)=44
2a²+a+b(2a+1)=44+a²+a
a(2a+1)+b(2a+1)=44+a²+a
(a+b)(2a+1)=44+a²+a;
but remember Q(a)=44+a²+a
then,
(a+b)(2a+1)=(2a+1)[(a/2)+(1/4)]+(175/4)
make 4 times the equation:
(4a+4b)(2a+1)=(2a+1)[2a+1]+175
(4a+4b)(2a+1)=(2a+1)²+175
(4a+b)(2a+1) - (2a+1)² =175
(2a+1)[(4a+4b)-(2a+1)]=175
(2a+1)[4a-2a+4b-1]=175
(2a+1)(2a+4b+1)=175
Where are the p and q? It is here the term p=(2a+1) and it the term q=[(2a+1)+4b].
■(2a+1)[(2a+1)+4b]=175; n=2a+1
better make like this way:
■n(n+4b)=175
Now, we can thing the possibilites:
n×(n+4b)=5²×7;
with
■D(175)={1;5;7;25;35;175}
And
■ n ⊂ D(175) (I)
then,
n ⊂ {1;5;7;25;35;175}
//////////////////////////////////////////////////
Considering the domain of the question:
remember that n=2a+1 and consider a ∈ N* because the question already said it. And, remember b ∈ N. And think about a domain of a and b.
So,
retake a²+2ab+b=44
you can see that each term of the equation cannot be bigger than 44. Then,
a²
Very nice solution! I really appreciate that ❤
I like your factorization solution, but I think this can be solved simply in this way(not sure is acceptable for math olympiads): since a>0,b>0 than a^2b=8,a=3=>b=5
Very nice! ❤
I would go in a similar way. Factorizing is a pain in the ass so I would just separate b
b=(-a^2+44)/(2a+1) |X2
2b=(-a(2a+1-1)+88)/(2a+1) |X2
4b+2a=(2a+1-1+176)/(2a+1)
4b+2a-1=175/(2a+1)
(4b+2a-1)(2a+1)=175 and from this we can look for all possible positive or negative pairs which can satisfy the equation
a^2+ab+ab+b=44
a(a+1)+b(a+1)=44
(a+1)(a+b)=44=4*11 или=2*22 или 1*44, если в целых. Подходит последнее: ответ: a=0, b=44.
1. а+1=4, a=3,
a+1)=4, (a+b)=11; ответ: a=3, b=8
Nice
2,8 just by putting a=2 and solving for b, if b comes to be a positive integer, its the correct answer
Very nice! ❤
4a^2+8ab+4b=176
4a^2-1+4b(2a+1)=175
(2a-1)(2a+1)+4b(2a+1)=175
Very Nice! ❤
a = 0 thì b =44. a khác 0 thì f(a) có nghiệm nguyên dương.
A^2+2AB+B=B（2A+1）+A^2-1/4+1/4=（A+1/2）（2B+A-1/2）+1/4=44，（2A+1）（2A+4B-1）=175
Very nice! ❤
HƯỚNG LÀ: A.B =0. Hằng đẳng thức. Hoặc phân tích đa thức. Đưa về A.B =0
44 is small. Just try A from 1 til 6.
Very nice! ❤
(a²+2ab+b²)-b²+b=44
(a+b)²-(b²-b+(1/2)²)+(1/2)²=44
(a+b)²-(b-1/2)²=44-1/4
(a+b+b-1/2)(a+b-b+1/2)=175/4
(a+2b-1/2)(a+1/2)×4=175
(2a+4b-1)(2a+1)=175=5*5*7*1
Case1 175*1
2a+1=1 ,a=0 (X)
Case2 35*5
2a+1=5 ,a=2 ,8+4b-1=35 ,b=7
Case3 25*7
2a+1=7 ,a=3 ,6+4b-1=25 ,b=5
(a,b) =(2,7) or =(3,5)
Very nice! ❤
Watching from Philippines sir
Thank you so much! ❤
First,dont use a line for division when you are multiplying!!This confuses students.In addition ,this i s much easier solved by trial and error.We know a is limited between 1 and 6.Just try them!!
Re-arrange, isolating b
b = (44 - a²)/(1 + 2a)
0 < a < 7
Noted! very nice suggession! ❤
a = -4, b = -4.
Yes, you are right! ❤
A.B=0. Làm như thế nào. 44 là chỗ khó. MỐI LIÊN HỆ a.b và 24.
44 (mod 12)=8 b=8😊
Very nice! ❤
@SALogics DEGERLİ dostum tüm üslü sayılar mod la çözülebilir.Bunun çalışmasını yapıyorum .Aslında asal sayılarla ilgilenirken üslü sayılar tesadüfen karşıma çıktı ve logaritma limit trigonometri 10 yaşındaki çocugunn çözebileceği yöntemler buldum.😁😁Örneğin x^3+y^3+z^3=a^3 mü sorusu cevabı irrasyonel olarak vardır.
a shall > 0
Yes
(A,B):(2,8)
Very nice! ❤
Case 1. a=0, but a>0. Nothing
If the original equation is transformed into an equation that has the one variable on one side of the equation and the other variable on the opposite side of the equation, we will have: b=(44-a^2)/(2a+1).
From the above relationship, if we assign any value for "a", we will obtain a corresponding value for "b". Therefore, there are infinitely many pair of solutions for "a" and "b". Examples are: (a,b)=(0,44), (2,8), (3,5), (12,-4), (17,-7), (87,-43), (-1,-43),(-3,-7), (-4,-4), (-13,5), (-18,8), (-88,44) and so on.
Very nice trick! ❤
يجب أن تلتزم بأن تكون قيمة كل من a>0 and b
Okay, that's nice. Thank you for the input; however, there's only a finite number of integral solutions.
@@ronbannon Thanks. I agree with you that there are only a finite number of integer solutions. However, the original problem is presented in a way that it simply asks for a=?, b=?. No restrictions are mentioned beforehand that a, b must be integer values. It is only when you click to play the video will you see the restrictions (a>0 and b>0). As a huge fan of these mathematical puzzles, I always try to solve the problems presented using my own analysis and approach. After I've done that, that's the time I play the video to verify if my answers are correct. Just to refresh you, notice that the original equation presented is actually an equation of a parabola (where the variables "a" and "b" are used instead of "x" and "y" found in analytic geometry textbooks. Since the parabola is a continuous locus of points, there would be an infinite number of points along its curve (unless you impose a restriction on the range and nature of values for "x" and "y".
@@ronbannon Thanks. I agree with you that there are only a finite number of integer solutions. However, the original problem is presented in a way that it simply asks for a=?, b=?. No restrictions are mentioned beforehand that a, b must be integer values. It is only when you click to play the video will you see the restrictions (a>0 and b>0). As a huge fan of these mathematical puzzles, I always try to solve the problems presented using my own analysis and approach. After I've done that, that's the time I play the video to verify if my answers are correct. Just to refresh you, notice that the original equation presented is actually an equation of a parabola (where the variables "a" and "b" are used instead of "x" and "y" found in analytic geometry textbooks. Since the parabola is a continuous locus of points, there would be an infinite number of points along its curve (unless you impose a restriction on the range and nature of values for "x" and "y".
Học Toán là Đoàn kết Gia Đình vì Hợp Tự nhiên . KHÁCH QUAN
Hãy tiếp tục học hỏi! ❤
Essa eu achei difícil
a prática leva à perfeição ❤
Tôi yêu Messi. Cậu ấy làm giáo Viên tốt
Messi là ai? ❤
Bạn không yêu Messi à? Cậu ấy là Giáo viên trên sân. Không phải cầu thủ đâu.
Thầy giống Messi quá.
Solution (0 ; b) n'est pas solution dans Z* (a > 0)
Oui, tu as raison ! ❤
@@SALogics😳🤣😡
a,b є N => a=1,2,3...
a,b є Z+ => a=0,1,2,3...
😬 The day before:
@habeebalbarghothy6320
Also (a,b)=(0,44) is possible.
@SALogics. ❤️
Yes, you are right!❤️
😂 Verify: 0²+2•0•44+44=44 !!!
The main reason u are not getting support even after such good content is that your channel name is salogic, but if someone searches sa logic, then your channel does not appear, so plz get a better name for your channel related to maths..For example Maths GO!
Nice suggession! ❤
Guessworks
Really? ❤
why it cannt be (1,44/3)
got it
because 44/3 is not an integer! ❤
Nợ mai tính tiếp
Brilliant!!!
Thank you so much! ❤
無聊