Question

Prove that there are no natural numbers, which are solutions of 15x^2 - 7y^2 =9

Answer 1

@funinabox

Answer 2

I would start by showing that x and y are both multiples of 3

Answer 3

ok , what to do after that ?

Answer 4

Re-arrange:\[\bf y^2=\frac{ 15x^2-9 }{ 7 }=\frac{ 3 }{ 7 } (5x^2-3)\]If the solutions are natural numbers, then they must be non-negative, hence:\[\bf y=+\frac{ \sqrt{3} }{ \sqrt{7} }\sqrt{5x^2-3}\]This can be re-written as:\[\bf \frac{ y }{ \sqrt{5x^2-3}}=\frac{ \sqrt{3} }{ \sqrt{7} }\]We must note that sqrt{3} and sqrt{7} are both irrational. This implies that 'y' must also be irrational. Comparing the denominators we get:\[\bf 5x^2-3=7 \implies x = \pm \sqrt{2}\], and sqrt{2} is also irrational.

Hence there is no natural number values of "x" and "y" which satisfy the given equation.

Answer 5

@goformit100

Answer 6

Thank you sir

Answer 7

@genius12

are you saying that both x and y have to be irrational to satisfy the original equation?