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# Algebra in number theory

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Find all integers m such that the quadratic equation x^2-mx+2m=0 has integer solutions.

So after some work i've got 4 ordered pairs
(8,2)
(4,4)

(-2,-8)

(-4,4)

So i don't know if i should subsitute here and if i should i don't know how to start subbing. Thanks in advance for any help.

Apr 10, 2023

#1
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We want to find all integers m such that the quadratic equation x^2 - mx + 2m = 0 has integer solutions.

Let's use the quadratic formula to solve for x:

x = (m ± √(m^2 - 8m)) / 2

For x to be an integer, we need the discriminant m^2 - 8m to be a perfect square, say k^2. That is:

m^2 - 8m = k^2

Rearranging, we get:

(m - 4)^2 - 16 = k^2

Adding 16 to both sides and factoring, we get:

(m - 4 - k)(m - 4 + k) = 16

Since m is an integer, the factors on the left-hand side must also be integers. The only possible pairs of factors of 16 are (1, 16), (-1, -16), (2, 8), (-2, -8), and (4, 4). So we have five cases to consider:

Case 1: m - 4 - k = 1 and m - 4 + k = 16 Adding the two equations, we get:

2m - 8 = 17 m = 25/2

This is not an integer, so there are no solutions in this case.

Case 2: m - 4 - k = -1 and m - 4 + k = -16 Adding the two equations, we get:

2m - 8 = -17 m = -9/2

This is not an integer, so there are no solutions in this case.

Case 3: m - 4 - k = 2 and m - 4 + k = 8 Adding the two equations, we get:

2m - 8 = 10 m = 9

Substituting m = 9 into the original quadratic equation, we get:

x^2 - 9x + 18 = 0

Factoring, we get:

(x - 3)(x - 6) = 0

So the integer solutions are x = 3 and x = 6.

Case 4: m - 4 - k = -2 and m - 4 + k = -8 Adding the two equations, we get:

2m - 8 = -6 m = 1

Substituting m = 1 into the original quadratic equation, we get:

x^2 - x + 2 = 0

This quadratic equation has no integer solutions.

Case 5: m - 4 - k = 4 and m - 4 + k = 4 Adding the two equations, we get:

2m - 8 = 8 m = 8

Substituting m = 8 into the original quadratic equation, we get:

x^2 - 8x + 16 = 0

Factoring, we get:

(x - 4)^2 = 0

So the only integer solution is x = 4.

Therefore, the only integers m that satisfy the given condition are m = 8 and m = 9.

Apr 10, 2023
#2
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I think this answer is really close but im pretty sure b isn't restricted to being a positive integer.

Apr 11, 2023
#3
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Well, I believe the answer is m=6 and m=11.

The angel number 1212 is commonly seen by "woke" people during their awakening.

We need to find all integers m such that m^2 - 8m is a perfect square. Let's complete the square by adding 16 to both sides:

m^2 - 8m + 16 = (m - 4)^2

____________________

So now we have:

m^2 - 8m + 16 - (m - 4)^2 = 0

(m - 4)^2 - (m^2 - 8m + 16) = 0

m^2 + 16m - 32 = 0

____________________

We can factor out -1 to get:

m^2 - 16m + 32 = 0

Using the quadratic formula, we get:

m = 8 ± 2sqrt(2)

_____________________

So the possible integer solutions for m are:

m = 6, 11

Substituting each value of m into the quadratic equation, we can check that they indeed produce integer solutions:

For m=6: x^2 - 6x + 12 = 0 --> x = 3 ± sqrt(3)

For m=11: x^2 - 11x + 22 = 0 --> x = 11 ± sqrt(33)

Again, I believe it is m=6 and m=11. Someone correct me if I am wrong! :-)

numberfreak  Apr 12, 2023