A number theory problem by Antonio Dottori
x + x y + y = 7 1 How many non-negative integer solutions (x,y) exist for this equation?
The answer is 12.
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1 solution
You could just use the fact that 7 2 = 2 3 × 3 2 , so 7 2 has ( 3 + 1 ) ( 2 + 1 ) = 1 2 positive divisors, so there's 1 2 pairs of positive integers x + 1 , y + 1 , or equivalently 1 2 , pairs of non-negative integers, x and y .
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That's true! I wrote them that way because my original problem was to find all the solutions, not the number of solutions =) Also, everyone who couldn't find an answer would know how to find them. Thanks a lot for the idea!
Done the same way...Great problem Antonio!!!!
x + x y + y = 7 1 .
x ( 1 + y ) + y = 7 1 .
x ( 1 + y ) + y + 1 = 7 1 + 1 .
( x + 1 ) ( y + 1 ) = 7 2 .
So x + 1 and y + 1 will be the divisors of 72.
x + 1 = 1 , 2 , 3 , 4 , 6 , 8 , 9 , 1 2 , 1 8 , 2 4 , 3 6 , 7 2
y + 1 = 1 , 2 , 3 , 4 , 6 , 8 , 9 , 1 2 , 1 8 , 2 4 , 3 6 , 7 2
Listing all the solutions:
x + 1 = 1 , x = 0 , y = 7 1
x + 1 = 2 , x = 1 , y = 3 5
x + 1 = 3 , x = 2 , y = 2 3
x + 1 = 4 , x = 3 , y = 1 7
x + 1 = 6 , x = 5 , y = 1 1
x + 1 = 8 , x = 7 , y = 8
x + 1 = 9 , x = 8 , y = 7
x + 1 = 1 2 , x = 1 1 , y = 5
x + 1 = 1 8 , x = 1 7 , y = 3
x + 1 = 2 4 , x = 2 3 , y = 2
x + 1 = 3 6 , x = 3 5 , y = 1
x + 1 = 7 2 , x = 7 1 , y = 0
Concluding, the solutions are: ( 0 , 7 1 ) ; ( 1 , 3 5 ) ; ( 2 , 2 3 ) ; ( 3 , 1 7 ) ; ( 5 , 1 1 ) ; ( 7 , 8 ) ; ( 8 , 7 ) ; ( 1 1 , 5 ) ; ( 1 7 , 3 ) ; ( 2 3 , 2 ) ; ( 3 5 , 1 ) ; ( 7 1 , 0 ) , which gives 1 2 solutions.