x2−3y2=1376 As we can see L.H.S. is definitely a multiple of 3 and in R.H.S. 1376 leaves a remainder of 2 when divided by 3. There are three possibilities for x in R.H.S: (i) If x is multiple of 3, so is x2, and R.H.S. will leave a remainder of 1 when divided by 3 (ii) If x is of the form 3m + 1, x2will be of the form 3n + 1 and R.H.S will leave a remainder of 2. m, n ∈ N (iii) If x is of the form 3m + 2, x2 will be of the form 3n + 1 and R.H.S. will leave a remainder of 2. m, n ∈ N So R.H.S. can never be a multiple of 3, while L.H.S. is always a multiple of 3. Hence no real solution exists.