Find all positive integral solutions of and to .

Remember, x and y are positive integers.

Source: mathcontest.olemiss.edu 9/21/2009

**SOLUTION
**

Since and are positive, must be positive in order to make the above product positive. This implies that .

Also, since 105 is odd, all three factors must be odd, which implies that is even.

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

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 |
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 |
x |
x + y |
x – y |
y(x + y)(x – y) |
||

7 |
8 |
15 |
1 |
7(15)(1)=105 |

There is a solution when and .

**Check
**

?

. YES.

**Answer**: .

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