Question

Find the natural numbers which differ by 3 and the sum of whose square is 117​

Answer 1

`\\\mathrm{Let\ the\ two\ natural\ numbers\ be\ }x\ \mathrm{and}\ y.\\\mathrm{First\ case:}\\x=y-3.......(1)\\\mathrm{Second\ case:}\\x^2+y^2=117\\\mathrm{or,\ }(y-3)^2+y^2=117\ \ \ \ \\\mathrm{or,\ }y^2-6y+9+y^2=117\ \ \ [\mathrm{From\ equation(1)}]\\\mathrm{or,\ }2y^2-6y-108=0\\\mathrm{or,\ }y^2-3y-54=0\\\mathrm{or,\ }y^2-9y+6y-54=0\\\mathrm{or,\ }y(y-9)+6(y-9)=0\\\mathrm{or,\ }(y-9)(y+6)=0\\\mathrm{i.e.\ }y=9\ \mathrm{or}\ -6`

`\\y=-6\ \mathrm{is\ not\ possible\ because\ }y\ \mathrm{is\ a\ natural\ number.}\\\mathrm{\therefore\ }y=9\\\mathrm{When\ }y=9,\ x=9-3=6\\\mathrm{So\ the\ two\ natural\ numbers\ are\ 9\ and\ 6.}`

Answer 2

Let's represent the two natural numbers as x and (x + 3) since they differ by 3. The sum of their squares can be expressed as follows:

x^2 + (x + 3)^2 = 117

Now, we can solve the equation:

x^2 + (x^2 + 6x + 9) = 117

Combine like terms:

2x^2 + 6x + 9 = 117

Subtract 117 from both sides:

2x^2 + 6x - 108 = 0

Now, we need to factor the quadratic equation:

2(x^2 + 3x - 54) = 0

Now, factor the quadratic expression inside the parenthesis:

2(x + 9)(x - 6) = 0

Set each factor to zero and solve for x:

x + 9 = 0 --> x = -9 (This solution is not a natural number, as it's negative)
x - 6 = 0 --> x = 6

So, the two natural numbers that differ by 3 and whose sum of squares is 117 are 6 and 9.