Question

If ${{x}^{2}}-5{{y}^{2}}=1232$ , then how many pairs of (x, y) will satisfy the equation.

Answer 1

Hint: Now we are given with the equation ${{x}^{2}}-5{{y}^{2}}=1232$ . Now first note that there are 2 in units placed in RHS. This means the value of ${{x}^{2}}-5{{y}^{2}}$ will have 2 in its units place.
Now if we see $5{{y}^{2}}$ will always have 5 or 0 in its units place. This means in order to get 2 in units place we will need 2 or 7 in units place of ${{x}^{2}}$ . We will try to find x satisfying such conditions if any.

Complete step-by-step answer:
Now first let us consider the equation ${{x}^{2}}-5{{y}^{2}}=1232$ .
Now first of all this is one equation in two variables.
Since number of equations < number of variables it is impossible to actually solve and get the values.
Hence we will try to approach the problem with a different method.
Now first note that there are 2 in units placed in RHS. This means the value of ${{x}^{2}}-5{{y}^{2}}$ will have 2 in its units place.
Now consider $5{{y}^{2}}$ here ${{y}^{2}}>0$
Now let us see some positive multiples of 5.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50….
Hence we can say that multiples of 5 will always have 0 in its units place
Now consider ${{x}^{2}}$ now this number is such that when we subtract $5{{y}^{2}}$ from this we get a number with 2 in its units place.
Let us try to see what is happening here since we have $5{{y}^{2}}$ always has 5 or 10 in units place.
Let us say ${{x}^{2}}$ has a in its units place…………….. (1)
Then we have something like $a-5$ or $a-0$ in units place of ${{x}^{2}}-5{{y}^{2}}$ .
Now there are 2 units placed in RHS.
Hence ${{x}^{2}}-5{{y}^{2}}$ will also have 2 in its units place.
This means $a-5=2$ or $a-0=2$
Which gives us nothing but $a=7$ or $a=2$
Now from equation (1) we have that ${{x}^{2}}$ has a in its units place
Now let us check and find x for which this might be true
$\begin{align}
  & {{1}^{2}}=1 \\
 & {{2}^{2}}=4 \\
 & {{3}^{2}}=9 \\
 & {{4}^{2}}=16 \\
 & {{5}^{2}}=25 \\
 & {{6}^{2}}=36 \\
 & {{7}^{2}}=49 \\
 & {{8}^{2}}=64 \\
 & {{9}^{2}}=81 \\
 & {{10}^{2}}=100 \\
\end{align}$
Hence we can see that for no number we will ever get 2 and 7 in its units place.
Hence no values of x and y will satisfy the equation.

Note: Note that since ${{1}^{2}}=1$ all other numbers 11, 21, 31, 41… will have 1 in units place similarly all numbers 12, 22, 32, 42…. Will have 4 in units place and 13, 23, 33, 43 will have 9 in its units place. Hence we need not check squares of too many numbers just 1 to 10 will suffice to make this result