Square or Cube

Find all positive integral solutions of x and y to x^2y-y^3=105.
Remember, x and y are positive integers.
Source: mathcontest.olemiss.edu 9/21/2009

SOLUTION
x^2y-y^3=105
y\left (x^2-y^2\right )=105
y\left (x+y\right )\left (x-y\right )=105
y\left (x+y\right )\left (x-y\right )=3\times 5\times 7
Since x and y are positive, \left (x-y\right ) must be positive in order to make the above product positive. This implies that x>y.

Also, since 105 is odd, all three factors y, \left (x+y\right ), \textup{ and }\left (x-y\right ) must be odd, which implies that x is even.

So let’s try the three possible values for y: 3, 5, or 7.

\mathbf{y=3}

y

x

x + y

x – y

y(x + y)(x – y)

   

3

4

7

1

3(7)(1)=21

   
 

6

9

3

3(9)(3)=81

   
 

8

11

5

3(11)(5)=165

>105

 

Thus, there is no solution when y=3.

\mathbf{y=5}

y

x

x + y

x – y

y(x + y)(x – y)

   

5

6

11

1

5(11)(1)=55

   
 

8

13

3

5(13)(3)=195

>105

 

Thus, there is no solution when y=5.

\mathbf{y=7}

y

x

x + y

x – y

y(x + y)(x – y)

   

7

8

15

1

7(15)(1)=105

   

There is a solution when y=7 and x=8.

Check
x^2y-y^3=105
8^2\times 7-7^3=105?
64\times 7-343=105
448-343=105
105=105. YES.

Answer: x=8,\;y=7.

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About mvtrinh

Retired high school math teacher.
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