1. **Problem statement:** Prove that for any positive integers $m$ and $n$, the product $$(36m + n)(36n + m)$$ can never be a power of 2. 2. **Recall the definition:** A power of 2 is a number of the form $$2^k$$ where $k$ is a nonnegative integer. 3. **Key observation:** If $$(36m + n)(36n + m)$$ is a power of 2, then both factors must be powers of 2 themselves or one must be 1, because the prime factorization of a power of 2 contains only the prime 2. 4. **Check if either factor can be 1:** - Suppose $$36m + n = 1$$. Since $m,n$ are positive integers, $$36m \\geq 36$$, so $$36m + n \\geq 37$$, contradiction. - Similarly, $$36n + m \\geq 37$$. 5. **Therefore, both factors are at least 37 and must be powers of 2 if their product is a power of 2. Powers of 2 greater or equal to 37 are 64, 128, 256, ... 6. **Check the parity of the factors:** - Since 36 is even, $$36m$$ and $$36n$$ are even. - Adding $n$ or $m$ (positive integers) to an even number results in: - Even if $n$ or $m$ is even - Odd if $n$ or $m$ is odd 7. **If both factors are powers of 2, they must be even (except 1, which is excluded). So both $$36m + n$$ and $$36n + m$$ must be even. This implies $m$ and $n$ are both even. 8. **If $m$ and $n$ are both even, write $m=2m'$, $n=2n'$. Then:** $$ (36m + n)(36n + m) = (36(2m') + 2n')(36(2n') + 2m') = 2(36m' + n') \, 2(36n' + m') = 4(36m' + n')(36n' + m') $$ 9. **This shows the product is divisible by 4, and the problem reduces to the same form with smaller positive integers $m', n'$. Repeating this argument infinitely would imply $m$ and $n$ are divisible by arbitrarily large powers of 2, which is impossible for positive integers. 10. **Hence, the product cannot be a pure power of 2.** **Final conclusion:** For any positive integers $m$ and $n$, the product $$(36m + n)(36n + m)$$ can never be a power of 2.