1.

Show that there are innitely many triples (x, y, z) of integers such that x3 + y4 = z31.

Answer»

Choose x = 24r and y = 23r. Then the left side is 212r+1. If we take z = 2k, then we get 212r+1 = 231k. Thus it is suffcient to prove that the equation 12r + 1 = 31k has innitely many solutions in integers. Observe that (12 x 18) + 1 = 31 x 7. If we choose r = 31l + 18 and k = 12l + 7, we get 

12(31l + 18) + 1 = 31(12l + 7); 

for all l. Choosing l ∈ N, we get innitely many r = 31l + 18 and k = 12l + 7 such that 12r + 1 = 31k. Going back we have innitely many (x, y, z) of integers satisfying the given equation.



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