28. If x and y are natural numbers, find the number pairs (x, y) for which x^2 - y^2 = 31.
pata nahi
To find the number pairs (x, y) for which x^2 - y^2 = 31, we can start by factoring the left side of the equation.
The left side expression x^2 - y^2 can be factored using the difference of squares formula (a^2 - b^2 = (a + b)(a - b)). Applying this formula, we have:
x^2 - y^2 = (x + y)(x - y)
Now we have the equation (x + y)(x - y) = 31. Since 31 is a prime number, the only way to factor it is as (1)(31) or (-1)(-31).
So, we have two equations to solve:
1) x + y = 31 and x - y = 1
2) x + y = -31 and x - y = -1
For the first equation, solving these equations simultaneously will give us the value of x and y:
Adding the two equations: (x + y) + (x - y) = 31 + 1
Simplifying: 2x = 32
Dividing by 2: x = 16
Substituting the value of x in the equation x + y = 31:
16 + y = 31
Subtracting 16 from both sides: y = 15
So, we have one solution for (x, y): (16, 15).
For the second equation, solving these equations simultaneously will give us the value of x and y:
Adding the two equations: (x + y) + (x - y) = -31 + -1
Simplifying: 2x = -32
Dividing by 2: x = -16
Substituting the value of x in the equation x + y = -31:
-16 + y = -31
Subtracting -16 from both sides: y = -15
So, we have another solution for (x, y): (-16, -15).
In conclusion, the number pairs (x, y) for which x^2 - y^2 = 31 are (16, 15) and (-16, -15).
well, (n+1)^2 - n^2 = 2n+1, so one pair will be when n=15, so
16^2 - 15^2 = 256-225 = 31
think of (n+3)^2 - n^2
and so on.
(why not possible for (n+2)^2 - n^2 ?)