Solving a Nice Diophantine Equation in Three Ways

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#DiophantineEquations #NumberTheoryProblems

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[chaparral]

if these two fractions add to 1, they must be both equal to 1/2. If one is smaller than 1/2 then the other has to be bigger than 1/2 which is not possible as long as the enumerator is 1. So a has to be -3 and B has to be 5. q.e.d.

[indusrealty]

3rd method is how I used when the saw the question and right away got the answer. I feel this is a trivial one

[seanfraser]

Multiply both sides of the equation by (a+5)(b-3) and get
a + b + 2 = ab -3a + 5b -15, or
ab - 4a + 4b -17 = 0.

Add 1 to both sides and factor, getting (a+4)(b-4) = 1.
Since a and b are integers, a+4 and b-4 are integers. The only way the product of two integers can be 1 is if they're both 1 or both -1. We get two potential solutions:
a = -3, b = 5
a = -5, b = 3

The second of these is extraneous (since it makes the denominators of the original fractions 0), so the only solution is (a, b) = (-3, 5).

[mxsjncv]

If you make it b = (4a+17) / (a+4) , a<> -4, -5 , b<>3, 4 and if you draw this "moved" hyperbulla [simple task. a=4 horizonal asymptotus. b=-4 vertical asymptotus...] then you see that you have very few posibilities to check.

[satrajitghosh]

Both a+5 and b-3 being integers, present probem requires partitioning of unity into reciprocals of two integers.
Trivially two halves makes unity
or 1/2 + 1/2 = 1.
or a = 2-5, b = 2+3

[professorpoke]

I know the answer is pretty obvious but check this proper solution.

integers ± integers => integers,
hence, a+5 & b-3 € int .
Lets call a+5 = A & b-3 = B, A&B are some integers.
(1/A) + (1/B) = 1

If A=1 it makes B= ∞, similarly
If B=1 it makes A= ∞.
Hence A, B ≠ 1
Also A, B ≠ 0 [that will make fractions = ∞]

case1: A>1
1/A ≤ 1/2
1/B must be ≥ 1/2
0< B ≤ 2
hence B = 2 [since B ≠1]
hence A = 2.

case 2: A < 0
A ≤ (-1)
(-1) ≤ (1/A) < 0
which makes
1 < (1/B) ≤ 2
which makes
1 > B ≥ 1/2
B is not an integer
hence no solution in this case.

Hence A = 2 & B = 2 is the only solution.
Substituting our original variables
a+5 = A = 2
a = (-3)
b-3 = B = 2
b = 5

Hence (a, b) = (-3, 2) is the only solution.

[nasrullahhusnan]

The solution can be obtained by 2 methods.
• Let x=a+5 and y=b-3 --> 1/x+1/y=1
Note that both x and y can't be 0
x+y=xy
Substract both sides by 1: x-xy-1+y=-1
x(1-y)-(1-y)=-1
(x-1)(y-1)=1 --> [(x-1), (y-1)]={(1, 1), (-1, -1)}
(x, y)=(2, 2) since x≠0, y≠0.
Thus (a, b)=(-3, 5)
• a and b are integers, so are x and y.
1/x+1/y=1 means that 1/x=1/y=½ as for any value of x and y not equal to 0, the numerator can't be any value other than 1. Thus x=y=2 --> (a, b)=(-3, 5)
Note that equation is symetrical
x(1-y)-(1-y)=

[stevenlitvintchouk]

Move the b fraction term to the right and take the reciprocal of both sides to get: a + 5 = (b-3) / (b-4) which means that a= (1 / (b - 4)) - 4. Because a is an integer, 1/ (b-4) must also be an integer. Obviously that can only be the case if b = 3 or b = 5. But b cannot equal 3 because the original equation would have a division by zero. That leaves only one solution, b = 5. And from the above, a = -3.

[bosorot]

Easy way to think is Inverse of integers can like 1/x have a set of { 1, 0.5 , 0.333, 0.25, 0.2, the only way to pick 2 number in the set and add them together to get 1 is only 0.5 + 0.5 . There is no other way .

[lilkuzi]

if we set a = b -8 the equation would be 2/b -3 = 1 from there b would be equal to 5 and a would be equal to -3

[Skank_and_Gutterboy]

Method #4: Solving for b directly yields b = 3 + 1/(1-(1/(a+5))). For b to be an integer,
1 - (1/(a+5)) has to be a fraction with a numerator that equals 1. This is self-evident because you have to take its inverse and add it to 3 to get b. Having a numerator other than 1 means that you're adding a fraction to 3 instead of an integer, so b would not be an integer and would not be a valid solution.

So now we've established that 1 - (1/(a+5)) = 1/int., where "int." denotes any integer.
1 - (1/(a+5)) = 1/int. can be manipulated using a common denominator to get
(a+4) / (a+5) = 1/int. Taking the inverse of both sides, you get (a+5) / (a+4) = int. The only way for (a+5)/(a+4) to yield an integer is if a+4 equals 1 or -1. Using a = -5 yields an integer (zero) but that causes a divide-by-zero problem in the original problem equation, so that's no good. So a+4 must equal 1. Therefore, a can only equal -3, so it follows that b can only equal 5. The only solution is: (a, b) = (-3, 5).

[damiennortier]

Set a+5 = x and b - 3 = y. Make a common denominator, you got (a+b)/ab = 1. It means that a+b = ab so a+b - ab = 0 and then, (a-b)^2 + ab = 0. Set (a+b)^2 = u and ab = t and got an easy equation

[mcwulf]

Or just say that the only values for x and y are (2, 2) because if x>2 then y<2 which is impossible as y=1 means x = infinity.

[youtuberbooster]

But sir..these values don't verify the base equation🙁

[Youtuber]

Or x divided by y and y diveded by x then x=y