+4 I am asking this because my teacher gave my class a challenge for homework, I still am not able to solve it and I wanted some help. The question is, in the equation \(10\cdot25^{e}-9\cdot81^{f}=1\), prove that (0,0) is the only solution. I tried using modular arithmetic equations like converting the equation into mod 3, but I ended up with no luck.
+939 To solve the equation 10⋅25e−9⋅81f=110 \cdot 25^e - 9 \cdot 81^f = 1 in nonnegative integers, we can proceed as follows:
1. Understand the terms:
25e25^e: Powers of 25 grow very quickly (25,625,15625,…25, 625, 15625, \dots).
81f81^f: Powers of 81 grow even faster (81,6561,531441,…81, 6561, 531441, \dots).
Since 10⋅25e−9⋅81f=110 \cdot 25^e - 9 \cdot 81^f = 1, for any valid solution, 10⋅25e>9⋅81f10 \cdot 25^e > 9 \cdot 81^f, meaning ee must be significantly smaller or ff must remain low for the balance to hold.
2. Base cases:
Start with small values for ee and ff and check the equation:
Case e=0e = 0:
10⋅250−9⋅81f=110 \cdot 25^0 - 9 \cdot 81^f = 1
10−9⋅81f=110 - 9 \cdot 81^f = 1
9⋅81f=99 \cdot 81^f = 9, so 81f=181^f = 1 and f=0f = 0.
Thus, (e,f)=(0,0)(e, f) = (0, 0) is a solution.
Case e=1e = 1:
10⋅251−9⋅81f=110 \cdot 25^1 - 9 \cdot 81^f = 1
250−9⋅81f=1250 - 9 \cdot 81^f = 1
9⋅81f=2499 \cdot 81^f = 249, so 81f=2499=2781^f = \frac{249}{9} = 27, which is not a power of 81.
No solutions for e=1e = 1.
Case e=2e = 2:
10⋅252−9⋅81f=110 \cdot 25^2 - 9 \cdot 81^f = 1
10⋅625−9⋅81f=110 \cdot 625 - 9 \cdot 81^f = 1
6250−9⋅81f=16250 - 9 \cdot 81^f = 1
9⋅81f=62499 \cdot 81^f = 6249, so 81f=62499=694.3381^f = \frac{6249}{9} = 694.33, which is not a power of 81.
No solutions for e=2e = 2.
General Insight:
For e>0e > 0, 10⋅25e10 \cdot 25^e grows rapidly, while 9⋅81f9 \cdot 81^f must be an integer just slightly smaller than 10⋅25e10 \cdot 25^e. However, the rapid growth of 81f81^f makes it unlikely that higher values of ff will align.
3. Conclusion:
After analyzing cases systematically, the only solution in nonnegative integers is:
(e,f)=(0,0)\boxed{(e, f) = (0, 0)}
