Dear Experts,
Can we solve this as below:
We are given that x^6−y^2=127...
or(x^3)^2 −y^2 = 127[in the normal A^2-B^2 format] .
Without knowing that 127 is a prime or not, we know that x & y can assume 4 values - i.e. -x or +x and -y or +y, yet the equation will remain same as both as squares and will always be +ive. Thus we get 4 different pairs of x & y. But the questions requires only +ive pairs, which is only one!
So ans is 1 pair.. Works? Any flaws? Thanks.
...
Can we solve this as below:
We are given that x^6−y^2=127...
or(x^3)^2 −y^2 = 127[in the normal A^2-B^2 format] .
Without knowing that 127 is a prime or not, we know that x & y can assume 4 values - i.e. -x or +x and -y or +y, yet the equation will remain same as both as squares and will always be +ive. Thus we get 4 different pairs of x & y. But the questions requires only +ive pairs, which is only one!
So ans is 1 pair.. Works? Any flaws? Thanks.
...



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