## Problems Based On Concept 1

254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is

a) prime

b) composite

c) prime square

d) a ,b,c

e) a,c

**255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such m’s**

**256) Find the number of pairs of positive numbers (x,y) such that x^3-y^3=100 and (x-y) and xy are integers.**

**a) 0 b) 1 c) 2 d) 4 e) none of these **

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Is the answer for 254 as

a) prime number

b) the m has 2 values and they are 10 and 55

TriptiAugust 27, 2008 at 10:04 am

no correct option is e)

outtimedAugust 27, 2008 at 10:32 am

Hi…

I could not solve the first problem.

255]

As it has 3 factors.. it is square of some prime number…

we get equation as (m+12+k) (m+12-k) = 125… so we get the factors 1,5,25,125… If we consider 25 (1*5*5)we get m = 3…

256]Answer is option 1… no pair will satisfy the value…

Can you help me with solutions?

RashmiAugust 27, 2008 at 4:11 pm

@Rashmi here are the solutions

255

for any natural number N=P^a.Q^b.R^c.. where P,Q,R are distinct prime numbers and

a,b,c are integers

no of factors is (a+1)(b+1)(c+1)

now as the no of factors is 3 one of a,b,c=2 and rest are zero

hence N is a prime square

let m^2+24m+21=p^2

then (m+12)^2+21-144=p^2

(m+12+p)(m+12-p)=123

m+12+p=123,41 m+12-p=1,3

2(m+12)=124,44

m=50,10

we check p=61,19 which are primes, hence both solution work

sum is 50+10=60

256)x^3-y^3=(x-y)[(x-y)^2+3xy]

let x-y=m and xy=n

then 100=m^3+3mn

clearly m<=4 and also m is a factor of 100

hence m=1,2,4

Case I when m=1

n=33

x-y=1 and xy=33

x^2 – x – 33 = 0

solving and taking positive value

x = (1+√(133))/2. Similarly we can find y.

case 2 m=2 n=46/3 not an integer

rejected

case 3 m=4 n=3

Since x – y = 4 we get x – 3/x = 4

Hence x^2 – 4x – 3 = 0

Solving we get x = (4 + √28)/2 = 2 + √7

hence we get two solutions

one for m=1 and other for m=4

hence two pairs

outtimedAugust 27, 2008 at 4:36 pm

254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is

a) prime

b) composite

c) prime square

d) a ,b,c

e) a,c

(x-y)(x+y) = a

Since a is an odd integer both the terms of the product chould be odd. Which is possible when 1 of the x/y is even and other is odd.

Since(x,y) has only one solution => that a can be written as product of 2 nos in only 1 way which is possible only when a is a prime number.

Assuming a to be positive. x>y

x-y = 1

x+y=p

x=(p+1)/2

y=(p-1)/2

I am not sure about prime square. But if it is square of prime no then a can be written as 1,p^2 or p,p

Here p,p is not possible as in that case y = 0 which is not possible as it is positive.

Is it correct, outtimed?

AmitAugust 28, 2008 at 6:36 pm

@ Amit

the correct answer is e) that is both prime and prime square !

for prime you have proved

for prime square see this

(x+y)(x-y)=p^2

so =(p^2+1)/2 and (p^-1)/2

outtimedAugust 28, 2008 at 7:25 pm

Great work, outtimed. Thank you very much for this blog.

Can we have more problems based on Concepts?

I think just trying to solve problems without having knowledge of concept doesn’t help much. Concepts followed by problems is really great idea. I will be following the blog regularly from now on. 🙂

AmitAugust 28, 2008 at 7:44 pm

@ Amit, yeah for sure I will be putting up more lessons, and every lesson will follow with more questions based on those lessons.

But there are also people who are good in concepts just need enough material to challenge them, so for them we have problems of the week and puzzles !

Thanks!

Have Fun !

outtimedAugust 28, 2008 at 7:49 pm

Hi outtimed ,

Dude!!! Thanks for the explanation …

Did some silly mistakes like 19 instead of 21 .. Phew…

Anyways…

Good Blog… Thanks to my friend who told me about it…

Keep good work!!!

RashmiAugust 29, 2008 at 11:16 am