practice problems

Q1. If x^n + y^n is a prime number for some x,y > 1 then n is necessarily:
Options: Power of 2, Multiple of 4, Multiple of 3, Odd

S1.
Answer: Power of 2.
Proof by contradiction:

Assume that n has an odd factor k > 1.
n = m.k
x^n + y^n = (x^m)^k + (y^m)^k
A^k + B^k = (A+B)(A^k-1 -A^k-2.B ....+ B^k-1) for odd 'k'.
Here A = x^m and B = y^m
x,y > 1 m>=1 => A+B >= 4
And the other factor is also more than 1 since A^k + B^k > A + B since k > 1.
So A^k + B^k is not a prime.
So n can't have any odd factor.
So n has to be power of 2.
H.P.

For e.g.
n = 2
x = 2, y = 3
4 + 9 = 13

Q2.
A natural number has last digit 8 in base 16 representation. What will be the remainder, when it is divided by 8 , in decimal system
S2.
0

Q3. If a prime can be written as sum of two perfect square, then the remainder when it is divided by 8 cannot be
S3.
Any square upon division by 8 will leave 0,1,4 as remainders.
So adding 2 of them can only give 0,1,2,4,5 as remainders.
And if the number is prime, remainder can't be 0,4.
So only possible values are 2,4,5.
And not possible values are 0,1,3,6,7.

Q4. The number of pairs of positive integers (m,n) such that m^4 + 4n^4 is prime is:
S4.

= (m^2)^2 + (2n^2)^2 + 4m^2.n^2 - 4m^2.n^2
= (m^2 + 2n^2)^2 - 4m^2.n^2
= (m^2 + 2n^2 + 2mn)(m^2 + 2n^2 - 2mn)

First factor is obviously >= 5
For it to be prime, second factor has to be 1.
m^2 + 2n^2 - 2mn = (m-n)^2 + n^2 = 1
=> (m-n)^2 = 1 - n^2
LHS is >= 0.
=> RHS = 0 or 1
=> n = 1,0
=> m = n = 1 is the only solution.
So only pair is (1,1).

Q5. Let x be a natural number < 10. 32x in base y is 263 in base 10. Find x + y.
S5.
3y^2 + 2y + x = 263
=> x =263 - 3y^2 - 2y
Since x < 10 we can try few values of y.
y = 8 gives x = 55 and y = 9 gives x = 2. y=10 makes x negative so we stop here.
So answer 2 + 9 = 11.


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