Question

Find the smallest solution in positive integers of $x^2-14y^2=1$.

Answer 1

Hint: In this particular question use the concept of hit and trial method by substituting y = 1, 2, 3.... in the given equation until we get the smallest positive integer for x so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given equation
$x^2-14y^2=1$.................. (1)
Now we have to find the smallest solution of the above equation in positive integers.
So the simplest method to solve this is by hit and trial method.
As we need positive integers, so we cannot substitute y = 0.
So start substituting, y = 1, 2, 3,..... in equation (1) until we get the smallest positive integer for x.
So when, y = 1
So from equation (1) we have,
$\Rightarrow x^2-14{\left(1\right)}^2=1$
Now simplify it we have,
$\Rightarrow x^2=1+14=15$
Now take square root on both sides we have,
$\Rightarrow x=\sqrt{15}$
So square root of 15 is not an integer value so it is not the required solution.
Now when, y = 2
So from equation (1) we have,
$\Rightarrow x^2-14{\left(2\right)}^2=1$
Now simplify it we have,
$\Rightarrow x^2=1+14\left(4\right)=57$
Now take square root on both sides we have,
$\Rightarrow x=\sqrt{57}$
So square root of 57 is not an integer value so it is also not the required solution.
Now when, y = 3
So from equation (1) we have,
$\Rightarrow x^2-14{\left(3\right)}^2=1$
Now simplify it we have,
$\Rightarrow x^2=1+14\left(9\right)=127$
Now take square root on both sides we have,
$\Rightarrow x=\sqrt{127}$
So square root of 127 is not an integer value so it is also not the required solution.
Now when, y = 4
So from equation (1) we have,
$\Rightarrow x^2-14{\left(4\right)}^2=1$
Now simplify it we have,
$\Rightarrow x^2=1+14\left(16\right)=225$
Now take square root on both sides we have,
$\Rightarrow x=\sqrt{225}=\pm 15$
So square root of 225 is an integer value so it is the required solution.
So the positive value of x when y = 4, is 15
So the required smallest solution in positive integers is (15, 4).
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that 0 is neither consider as positive nor negative, and the basis to solve these types of problem quickly is by hit and trial method, so substitute the values of y starting from 1 in the given equation and solve for x as above we will get the required smallest solution in positive integers.