Problem: If 2a+2b=3c+3d, the number of integers a,b,c,d which can possibly be negative, is, at most:
Answer Choices:
A. 4
B. 3
C. 2
D. 1
E. 0
Solution:
Suppose one of a,b,c,d is a negative integer, say d; then, we have 2a+2b−3c=3−d, an impossibility since the left side is an integer while the right side is a non-integer. Similar reasoning shows that no two, and no three, of the exponents can be negative integers.
If all four are negative integers, then we may write 2a1​+2b 1​=3c1​+3d1​, with a,b,c,d positive integers. Then 22⋅2b2b+2a​=3c⋅3d3d+3c​
If a<b, divide the numerator and the denominator of the left fraction by 2a, and obtain 2b(=D1​)2b−1+1(=N1​)​=3c⋅3d(=D2​)3c+3d(=N2​)​, or, equivalently, N1​D2​=N2​D1​. But N1​ is odd, D2​ is odd, N2​D2​ is odd, N2​ is even, D1​ is even, N2​D1​ is even-a contradiction.
If b<a. use instead the divisor 2, and prove that the same contradiction arises.
If b=a, the left fraction can be reduced to 2a−11​, and, again, the same contradiction arises.