將寫成傅里葉積分形式于是有e^(ik 13/42)+e^(ik )=e^(ik 1/6)+e^(ik 1/7)兩邊計算各自復(fù)共軛,并令k/42=se^(i13s)+e^(i42s)-e^(i6s)-e^（i7s)=0e^(i7s)+e^(i36s)-1-e^（is)=0（e^(i6s)-1）{e^(i30s)+e^(i24s)+e^(i18s...用到了，因為只有有界函數(shù)其傅立葉變換系數(shù)才是有界的。任何有界函數(shù)可以寫為f(x)=integrate[g(k)e^ikx,{k,-ifinity,ifinity}]上面等式為integrate[g(k)e^ik(x+13/43),{k,-ifinity,ifinity}]+integrate[g(k)e^ik(x+1),{k,-ifinity,ifinity}]=integrate[g(k)e^ik(x+1/6),{k,-ifinity,ifinity}]+integrate[g(k)e^ik(x+1/7),{k,-ifinity,ifinity}]integrate[g(k)e^ikx{e^(ik 13/42)+e^(ik )-e^(ik 1/6)-e^(ik 1/7)},{k,-ifinity,ifinity}]=0g(k)有界，要求{e^(ik 13/42)+e^(ik )-e^(ik 1/6)-e^(ik 1/7)}=0可以計算k，而且只有幾個k函數(shù)變?yōu)閒(x)=Sum[g(k)e^ikx,{k,k1,k2,..}]如果只有一個kf(x)=g(k)e^ikx+g*(k)e^(-ikx)這個函數(shù)的周期為 k/2Pi n用平移算子 T=e^(d/dx)f(x+13/42)+f(x)=f(x+1/6)+f(x+1/7) 變?yōu)?T^(13/42)f(x)+f(x)=T^(1/6)f(x)+T^(1/7)f(x){T^(13/43)+1-T^(1/6)-T^(1/7)}f(x)=0(T^(30/42)+T^(24/42)+T^(18/42)+T^(12/42)+T^(1/42)+1)(T^(1/7)-1)f(x)=0(T^(30/42)+T^(24/42)+T^(18/42)+T^(12/42)+T^(1/42)+1)(f(x+1/7)-f(x))=0如果算子(T^(30/42)+T^(24/42)+T^(18/42)+T^(12/42)+T^(1/42)+1)可逆（我不知道還能否因式分解），有(f(x+1/7)-f(x))=0