Question:
If $a$ and $ b$ are positive integers such that $a^2 - b^2$ is a prime number, then $a^2 - b^2$ is
If $a$ and $ b$ are positive integers such that $a^2 - b^2$ is a prime number, then $a^2 - b^2$ is
Updated On: May 11, 2024
- $a+b$
- $a - b$
- $ab$
- 1
Hide Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
$a $ and $ b$ are positive integers and $a^2 - b^2$ is a prime number.
Since, $a^2 - b^2 = (a+ b) (a - b)$ product of two numbers.
$\therefore$ Either $a + b = 1$ or $a - b = 1$
Case I : $a + b = 1 \Rightarrow $ Either $a = 0$ or $b = 0$ then $a^2 - b^2 = 1$ or $-1$
which is not a prime number.
$ \therefore$ This case is not possible
Case II : $a - b = 1 \rightarrow a$ and $b$ can be taken anything with $b < a$.
ln this case $a^2 - b^2 = a + b$ is a prime number.
$\therefore \:\:\: a^2 - b^2 = a + b.$
Since, $a^2 - b^2 = (a+ b) (a - b)$ product of two numbers.
$\therefore$ Either $a + b = 1$ or $a - b = 1$
Case I : $a + b = 1 \Rightarrow $ Either $a = 0$ or $b = 0$ then $a^2 - b^2 = 1$ or $-1$
which is not a prime number.
$ \therefore$ This case is not possible
Case II : $a - b = 1 \rightarrow a$ and $b$ can be taken anything with $b < a$.
ln this case $a^2 - b^2 = a + b$ is a prime number.
$\therefore \:\:\: a^2 - b^2 = a + b.$
Was this answer helpful?
0
0
Top Questions on Binomial theorem
- The sum of the coefficients of \( x^{2/3} \) and \( x^{-2/5} \) in the binomial expansion of \[ \left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9 \] is:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $m$ and $n$ be the coefficients of the seventh and thirteenth terms respectively in the expansion of \[\left( \frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2x^{\frac{2}{3}}} \right)^{18}.\]Then \[\left(\frac{n}{m}\right)^{\frac{1}{3}}\]is:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- If the term independent of \(x\) in the expansion of \[ \left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10} \] is 105, then \(a^2\) is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $\alpha = \sum_{r=0}^n (4r^2 + 2r + 1) \binom{n}{r}$ and $\beta = \left( \sum_{r=0}^n \frac{\binom{n}{r}}{r+1} \right) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$, then the value of $n$ is _____.
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $0 \leq r \leq n$. If \[^nC_{r+1} : ^nC_r : ^nC_{r-1} = 55 : 35 : 21,\]then $2n + 5r$ is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
View More Questions
Questions Asked in COMEDK UGET exam
- A $2\,kg$ mass lying on a table is displaced in the horizontal direction through $50\,cm$. The work done by the normal reaction will be
- COMEDK UGET - 2015
- work, energy and power
- The numbers of $ {H^+}$ ions present in 1 mL of a solution whose pH is 13
- COMEDK UGET - 2015
- Equilibrium
- A mixture of methylamine and dimethylamine is given to you. The reagents used to separate the components of the mixture are
- $\cot^{-1} (21) + \cot^{-1} (13) + \cot^{-1} (-8) =$
- COMEDK UGET - 2015
- Inverse Trigonometric Functions
- Pick the incorrect statement among those given below.
- COMEDK UGET - 2015
- Chemical bonding and molecular structure
View More Questions
Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.