A Fun Diophantine Equation (Not From Harvard 😜)
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- Опубліковано 7 лют 2025
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a^2+2ab+b=44
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![[Math]Find two solutions of the equation. Give your answers in degrees (0° ? ? 360°) and radians](http://i.ytimg.com/vi/eTwNlSGqhhs/mqdefault.jpg)
The following method is easy to remember and works best if a and b are restricted, esp if a, b, are integers >/= 0...
Isolate b so that b = (44 - a²)/(1+2a)
Set >/= 0 and solve the inequality so that a < 6
Now solve for a = 6, 5, 4, 3, 2, 1 . . . making a table is the best way
I got (3, 5) and (2, 8)
I got a merhod for mortals. If you do not see the trick to factorising just solve it
a^2+2ab+b=44
b(2a+1)=44-a^2
b=(-a^2+44)/(2a+1) do long division or factor out
b=(-a^2-1/2a+1/2a+1/4-1/4+44)/(2a+1)
b=-1/2a+1/4+175/(4*(2a+1)) multiply by 4
4b=-(2a-1)+175/(2a+1)
4b+2a-1=175/(2a+1)
And go from there
A slightly different method.
a^2 + 2ab + b = 44
a^2 + b(2a+1) = 44
// 2a+1 is a factor of 4a^2-1 (difference of two squares)
// multiply by 4 and subtract 1
(4a^2-1) + 4b(2a+1) = 175
// extract common factor
(2a+1)(2a + 4b - 1) = 175
// sum of factors = 4a + 4b
// let the factors be p & q
// find a+b directly
a+b = (p+q)/4
The rest follows...
Here's how I proceeded:
largest square number smaller than or equal to 44 is 36 whoose square root is 6 so
let a= 6
a²+2ab +b = 44
36 +12b +b = 44
13b = 8
b = 8/13
(a,b) = (6,8/13)
a+b = 86/13
You can let a = any number less than 6
example
let a= 2
a²+2ab+b = 44
4 +4b+b = 44
5b = 40
b= 8
(a,b) = (2,8)
a+b = 10
a^2 + 2ab + b = 44
(a + b)^2 = b^2 - b + 44
(a + b)^2 = (b - 1/2)^2 - 1/4 + 176/4
(a + b)^2 = (b - 1/2)^2 + 175/4
4(a + b)^2 = (2b - 1)^2 + 175
(2a + 2b)^2 - (2b - 1)^2 = 175
(2a + 1)(2a + 4b - 1) = 5*5*7
a + b = [±44; ±10; ±8]
a little more explanation
let c=2a+1
d=2a+4b-1
c+d=4a+4b
(c+d)/4 = a+b
(c,d) = (1,25×7) or (5,5×7) or (7,25)
so c+d= 176 or 40 or 32
(c+d)/4= +-44 or +-10 or +-8
(c+d)/4=a+b
Answer
a+b=+-44 or +-10 or +-8
check
a=(c-1)/2
c=+-1, +-5, +-7, +-25, +-35, +- 175
a=0;-1, 2;-3, 3;-4, 12,-13, 17;-18, 87;-88
b=44;-43,
@@davidseed2939 Nicely done. I did it the hard way and got the same answers from 12 factorisations.
@davidseed2939 c,d can be 12 different pairs. I think you have them all covered but it looks like just 3 in one line.
That's so cool!
Good! I also solved by this method.
(a + b)² = 44 - b + b²
Since Left Hand Side (LHS) is a perfect square, it follows RHS must be too, and possibly equal to 0, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144; i.e. k²
Integer solution for b is required for
ķ² - 44 = b(b-1) or product of two consecutive numbers, by inspection we see b = 5, 8..
Thus, (a + b) = 8, 10.. are solutions
a+b=[(2a+1)+(2a+4b-1)]/4 => a+b= (sum of factors)/4= +-8, +-10, +-44
[Where factors are (+-1,+-175), (+-5,+-35), (+-7,+-25)]
Or re-arrange to 4(a + b)² - (2b - 1)² = 175 or x² - y² = 175 where x = 2(a + b) and y = (2b - 1)
(x + y)(x - y) = 175 and check for 6 cases: 1•175, 5•35, 7•25, 25•7, 35•5, 175•1 etc
very nice
Another commenter worked out (as I did) that you can express b as a function of a thus: b = (44 - a*a)/(2a + 1)... Hence you can work out that a + b = (44 + a*a + a)/(2a + 1). The integer solutions with (a,b) equal to either (2,8) or (3,5) satisfy this formula for a+b .
a = 3, b = 5
9+30+5=44
Answer: 8
I mean, how come common students know the factor of those from those non existing value??? Yeah, assuming they solve lots problem, it's still not explaining how come they know they should multiply first, adding +K-K, and so on. I can make sure of it since not any algebra textbooks have this kind of solution
If only the first equation had b^2 instead b.
I thought it was until I read your message!
The thing about Diophantine equations is they change one thing to the equation, to make it diophantine. The equation given, might have infinitely many solutions unless you state the solutions must be integers
3 works. a = 3 and b = 5, for example.
Did you say positive integers?
I plugged in 4 for a and 8 for b, so a+b=4+8=12.
Are you crazy 🤣 how can you put a as 4 and b as 8 it is not satisfying
a=2@@MaheshKumar-lx1ku
Logarithmic inequalities, please.
😂
a+b=5+4=9 or
a+b=2+8=10.......May be
Explain later...
It is surprising how many really bad maths videos there are on YT.
Here is a systematic approach that is quick short and easy.
See that a must be less than 7.
Try a=6 and you get 13b = 8. No good
Try a = 5 and you get 11b = 19. No good
Try a = 4 and you get 9b = 28. No good
Try a=3 and you get 7b = 35 so b = 5.
Do you consider this a bad video?
Do you think your solution is a good one? 🤔
how you know a must be less than 7??
if a>7 then a^2>49