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Integers — MATHS STD 6 — Question

Gujarat BoardEnglish MediumSTD 6MATHSIntegers1 Mark
MCQ
Let $n$ be an integer number. Prove that the equation has infinitely $x^2 + y^2 = n + z^2$ many integer solutions.
  • ✓
    $z = x − 1$
  • B
    $z = x + 1$
  • C
    $z = x × 1$
  • D
    $z = x /1$
✓

Answer

Correct option: A.
$z = x − 1$
The equation is equivalent to $(x - z) (x + z) + y^2= n.$ If we set $x - z =$ then we obtain $2x - 1 + y^2 = n$ and $\text{x} = \frac{\text{n + 1 - y}^2}{2}$ . Now, it suffices tso take $y = n + m,$ where m is an odd integer to insure that $x$ is an integer as well. Indeed, if $m = 1$ kt $1$ then,
$\frac{\text{n + 1 - y}^2}{2} = \frac{\text{n+1-(n+2k+1)}^2}{2}=-\frac{\text{n (n + 1)}}{2}\\-2\text{nk}-2\text{nk}^2-2\text{k,}$
it is also an integer number. obviously an integer. Since $z = x - 1 z = x - 1$

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