Number of solutions of equation x ^2+ y ^2= a ^2+ b ^2+ c ^2 . where x , y , a , b , c are all prime

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4-2.Quadratic Equations and Inequations

normal

The number of solutions of the equation $x ^2+ y ^2= a ^2+ b ^2+ c ^2$. where $x , y , a , b , c$ are all prime numbers, is

$0$

$1$

more than $1$ but finite

infinite

(KVPY-2021)

Solution

(a)

Any prime other than 2,3 is of the form $6 k \pm 1$

$\therefore 6 k \pm 1= p$

$6 \lambda+1= p ^2$

$(I)$ If $2 \& 3$ are not solutions

$x^2 \equiv 1(\bmod 6)$

$y ^2 \equiv 1(\bmod 6)$

LHS $\equiv 2(\bmod 6)$

$RHS \equiv 3(\bmod 6)$

$(II)$ If $2$ is not then all are odd

$LHS$ $\equiv$ Even

$RHS$ $\equiv$ Odd

$C – I$ when $x = y =2$

$a^2+b^2+2=8$

Let $a =2$ then $b ^2+ c ^2=4$

which is not possible for prime ' $b$ ' \& 'c'

$C-II$ $x =2 ; y \neq 2$

$4+ y ^2= a ^2+ b ^2+ c ^2$

Not possible

$C-III$ $x \neq 2 ; y \neq 2$

Not possible

Standard 11

Mathematics

The polynomial equation $x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$ has

normal

(KVPY-2016)

Below are four equations in $x$. Assume that $0 < r < 4$. Which of the following equations has the largest solution for $x$ ?

normal

(KVPY-2011)

If graph of $y = ax^2 -bx + c$ is following, then sign of $a$, $b$, $c$ are

normal

Let $p, q$ and $r$ be real numbers $(p \ne q,r \ne 0),$ such that the roots of the equation $\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to .

hard

(JEE MAIN-2018)

The solution of the equation $2{x^2} + 3x – 9 \le 0$ is given by

medium

The number of solutions of the equation $x ^2+ y ^2= a ^2+ b ^2+ c ^2$. where $x , y , a , b , c$ are all prime numbers, is

Solution

Similar Questions

The polynomial equation $x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$ has

Below are four equations in $x$. Assume that $0 < r < 4$. Which of the following equations has the largest solution for $x$ ?

If graph of $y = ax^2 -bx + c$ is following, then sign of $a$, $b$, $c$ are

Let $p, q$ and $r$ be real numbers $(p \ne q,r \ne 0),$ such that the roots of the equation $\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to .

The solution of the equation $2{x^2} + 3x – 9 \le 0$ is given by

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