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, N1D2=N2D1. But N1 is odd, D2 is odd, N2D2 is odd, N2 is even, D1 is even, N2D1 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.