Problem:
Find the number of pairs (m,n) of positive integers with 1≤m<n≤30 such that there exists a real number x satisfying
sin(mx)+sin(nx)=2
Solution:
Note that the maximum of sin(mx) is 1 and is achieved when x=m360∘k+90∘ for any integer k. If sin(mx)+sin(nx)=2, then there exists a real number x such that sin(mx)=sin(nx)=1. Thus x=m1(360∘k+90∘)=n1(360∘ℓ+90∘) for some integers k and ℓ. Hence
m4k+1=n4ℓ+1
which is equivalent to (4k+1)⋅n=(4ℓ+1)⋅m. Because 4k+1 and 4ℓ+1 are odd, the greatest power of 2 dividing m must be equal to the greatest power of 2 dividing n. Let m=2t⋅m′ and n=2t⋅n′, where m′ and n′ are both odd. Then (4k+1)⋅n′=(4ℓ+1)⋅m′ and 4∣(m′−n′).
Conversely, if m′ and n′ are odd positive integers satisfying m′≡n′(mod4), then there exist positive integers k and ℓ such that the above equation holds:
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If m′ and n′ are congruent to 1 modulo 4, then setting 4k+1=m′ and 4ℓ+1=n′ leads to integer values for k and ℓ.
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If m′ and n′ are congruent to 3 modulo 4, then setting 4k+1=3m′ and 4ℓ+1=3n′ leads to integer values for k and ℓ.
Therefore the required integers k and ℓ exist if and only if m′ and n′ are both odd and 4∣(m′−n′), which means that either m′,n′∈{1,5,9,…,29} or m′,n′∈{3,7,11,…,27}. In the first case, m and n are distinct integers from {1,5,9,…,29}, from {2,10,18,26}, or from {4,20}. In the second case, m and n are distinct integers from {3,7,11,…,27}, from {6,14,22,30}, or from {12,28}. Hence there are
(82)+(42)+(22)+(72)+(42)+(22)=63
ordered pairs with the required properties.