2025/014) Show that there are infinitely many positive integers which cannot be expressed as the sum of squares.

we have $n^2 \equiv 0/1 \pmod 4$

So $m^2 + n^2 \equiv 0/1/2 \mod 4$

So any number of  of the form 4k + 3 cannot be expressed as sum of 2 squares

There are infinitely many of them 

Hence proved

2025/013) Prove that for every $n \in N$ the following proposition holds: $7|3^n + n^ 3$ if and only if $7 |3^nn^ 3 + 1$

Now 7 cannot be multiple of 7 because in that case 7 cannot be factor of either of them

Because 7 is prime so using Fermat's little theorem we have  

$n^6-1=0$

or $n^6 \equiv 1 \pmod 7\cdots(1)$

Now let $7| 3^n + n^3$

So multiplying  by n^3 we get

 $7| 3^nn^3 + n^6$

Or  $7 | 3^n n^3+1$

Similaly we can prove the only if part 

2015/012) Solve in integers $2^x + 1 = y^2$

 We have $2^x = y^2- 1 = (y+1)(y-1)$

As product of y+1 and y-1 is a power of 2 so both are power of 2.

y+1 and y-1 one of them is divisible by and another by 2.

If y-1 is divisible by 4 then $y+1 = 4k+2( k \ge 1)$  for some k and it has an odd factor 2k+1. this is a contradiction as it should not have any factor other than 2

If y +1 is divisible by 4 then y+1 has to be 4 as any other factor will be power of 2 or odd or combination and this is contradiction

So y+1 = 4 and hence putting in given equation x = 3.

So we have $x=3,y=3$

 

2015/011) Prove that $\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \frac{1}{a_4} + \frac{1}{a_5} + \frac{1}{a_6} = 1$ then at least of of $a_1,a_2,a_3,a_4,a_5,a_6$ is even

 We have

$\frac{1}{a_1} + \frac{1}{a_2}  + \frac{1}{a_3}  + \frac{1}{a_4}  + \frac{1}{a_5}  + \frac{1}{a_6}=$

$\dfrac{a_2a_3a_4a_5a_6+a_1a_3a_4a_5a_6+a_1a_2a_4a_5a_6+a_1a_2a_3a_5a_6+a_2a_2a_3a_4a_6+a_2a_3a_3a_4a_5}{a_1a_2a_3a_4a_5a_6}$

let us assume that $a_1,a_2,a_3,a_4,a_5,a_6$ each is odd 

The numerator each term is odd(being product of 5 odd numbers) and there are even number of numbers so numerator is even and denominator is odd so the value cannot be 1.

So at least one of them has to be has to be even. 

2025/010) What should be the middle number if the L.C.M. of 3 consecutive even numbers is 4896?

 Let us factor $4896 = 48 * 102 = 2^5 * 3^2 * 17$

Because LCM is 4896 any one of the numbers should be multiple of $q^5=32$ one should be multiple of $17$ and one must be a multiple of $3^2=9$ and it must be remembered that  one number may satisfy one or more criteria. And no number can have a factor other than 2 , 3, of 147 7 and any power of these 3 more than as specified above

Let us try with $2^5$ or 32 say a

If 32 is there it has to be first number as 30 cannot there ( 5 is a factor)

So let us consider $32,34,36$. 34 is multiple of 17 and 36 is multiple of 9.

Le us take valid multiple of 32 that is $32 * 3$ 2 adjacent numbers are 94 and 98 not permissible as 94 has a factor  47 and 98 has factor 7

similarly we can rule out others

So the 3 numbers are $32,34,36$ and middle number is 34.

2025/009) Show that $(2+\sqrt{5} )^n + (2-\sqrt{5} )^n$ is integer for $n\in Z^+$

 We shall prove the same by induction

For n = 1 we have $(2+\sqrt{5} )^1 + (2-\sqrt{5} )^1 = (2+\sqrt{5} ) + (2-\sqrt{5} ) = 4 $

For n = 2 we have $(2+\sqrt{5} )^2 + (2-\sqrt{5} )^2 = (4+4\sqrt{5} + 5) + (4-4\sqrt{5} +5) = 18$

Let us assume that it is true for k upto n. We shall show that it is  true for k = n+1

below we shall keep power 1 for clarity.

Let $T_n = (2+\sqrt{5} )^n + (2-\sqrt{5} )^n$

We have $T_nT_1 = ((2+\sqrt{5} )^n + (2-\sqrt{5} )^n) ((2+\sqrt{5} )^1 + (2-\sqrt{5} )^1)$

$= (2+\sqrt{5} )^{n+1} + (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n + (2-\sqrt{5} )^{n+1}$

$= ((2+\sqrt{5} )^{n+1} +  (2-\sqrt{5} )^{n+1})+ (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n $

$=T_{n+1}+ (2+\sqrt{5} )^n * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1(2-\sqrt{5} )^n $ using definition

$=T_{n+1}+ (2+\sqrt{5} )^{n-1} *  (2+\sqrt{5} ) * (2-\sqrt{5} )^1 + (2-\sqrt{5} )^1 (2-\sqrt{5} )(2-\sqrt{5} )^{n-1} $ 

$=T_{n+1}+ (2+\sqrt{5} )^{n-1} *  (-1) -1 * (2-\sqrt{5} )^{n-1} $ using product

$=T_{n+1} -1 ((2+\sqrt{5} )^{n-1}+ (2-\sqrt{5} )^{n-1}$ using T

$=T_{n+1} - T_{n-1}$

Or $T_{n+1} = T_nT_1 + T_{n-1}$

or $T_{n+1} = 4 T_n + T_{n-1}$ using the value of $T_1$'

so if it is integer for value upto n then it is integer for k = n+1

Proved

 

2025/008) Find prime numbers x,y,z such that $7(x+y+z) = xyz$

Because $x,y,z$ are prime one of $x,y,z$ must be 7 so without loss of generality we assume $z = 7$

Putting in original equation we have

$7(7+x+y) = 7xy$

Or $xy - x -y -7=0$

Or $xy-x-y +1 = 8$

Or $x(y-1) - (y-1)=8$

Or $(x-1)(y-1)= 8= 1 * 8 = 2 * 4$

We assume $x \ge y$

this gives $x-1=8$ or x= 9 in which case it is not prime so not proper

Or $x-1=4,y=1=2$ giving $x = 5,y=3$ giving $x=5,y=3,z=7$ or any permutation of the same

2025/007) If 2n+1 and 3n+1 are both perfect squares, then how could we prove that 40|n?

For n this to be divisible by 40 we need to show that it is divisible by 8 and 5 as 40 = 8 * 5 and 8 and 5 are co-primes  

 First let us prove that 8 divides n

Clearly 2n+1 is odd so this has to be square of odd number say 2k+1

Now $(2k+1)^2 = 4k^2 + 4k +1=4k(k+1)+1 = 4m+1$ when $m= k(k+1)$

or $2n+1 = 8m+1$ or $n= 4m$

so n is even and hence 3n+1 is odd so  square of odd number say 2p+1

$3n+1 = (2p+1)^2 = 4p^2 + 4p +1 = 4p(p+1) +1$

or $3n = 4p(p+1)$ as p(p+1) is even it can be written as 2q

as $3n+1 = 8q+1$5

or $3n = 8q$

So 8 divides 3n and as 8 and 3 are copimes so 8 divides n 

now let us prove that 5 divides n

let us find what are the numbers that can be remainder of $x^2$ when divided by 5

we have x of the form $ 5q+k$ where k is one of $0,\pm 1,\pm 2$

$x^2= 25q^2 + 10kq +k^2$

$= 5(5q^2 + 2k) + k^2$

so remainder when $x^2$ divided by by5 is 0 or 1 or 4

now let us find n for which 2n+1 give a remainder 0 or 1 or 4 else it cannot be a square.

$n = 5k$ gives $2n + 1 = 10 k  + 1 = 5 * 2k + 1$ remainder is     1 which is a valid candidate

$n = 5k+ 1$ gives  $2n + 1 = 10 k  +3 = 5 * 2k +3$ remainder is 3 which is a not a valid candidate

$n = 5k+ 2$ gives  $2n + 1 = 10 k  +5 = 5 * (2k +1)$ remainder is 0 which is a valid candidate

 $n = 5k+ 3 gives  $2n + 1 = 10 k  +7 = 5 * (2k +1)+2 $ remainder is 2 which is a not a valid candidate

$n = 5k+ 4$ gives  $2n + 1 = 10 k  +9 = 5 * (2k +1)+ 4 $ remainder is 4 which is a valid candidate

So valid candidates for 2n+1  being perfect square is 5k, 5k+2, 5k+ 4

and we need to check for these one 3n+1 is valid

we can apply above procedure to see  the$ n= 5k$ 3n+1 leaves remainder 1 which is valid and for $n=5k+2$ remainder is 2 and for $n=5k+4$ remainder is 3 both of which are invalid

so n is multiple of 5.

as n is divisible by 5 and 8 so by 40

2025/006) Let n be the smallest positive integer such that n is divisible by 20, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$

 Because $n^2$ is a perfect cube so n is a perfect cube

Because $n^3$ is a perfect square so n is a perfect square

so n is a perfect $6^{th}$ power

as n is divisible by 20$2^2 * 5}

so it is of the form $2^x5^y$ where x and y are multiple of 6

so $n = 2^6 * 5^6 = 10^6= 1000000$

number of digits in $n$ = 7

2025/005) Consider the formation of integer, $x = 1555555\cdots 526$ . How to show that whatever amount of digit 5 is placed between $1 \& 2 , 7 ∣ x$

We have

$x = 1555555\cdots 526$ number of 5's >=0$

so $x + 29 = 1555555\cdots 5$

Let it be a n digit number 

so $x + 29 = 10^{n-1} +  5 *111\cdots 1$                                  n-1 1's

or  $x + 29 = 10^{n-1} + \frac{5}{9}(10^{n-1} -1 )$

as 7 is not a factor of 9 

So we have $9(x+29) = 9 * 10^{n-1} + 5 * 10^{n-1} -5$

or $9x + 9 * 29 -5 = 14 * 10^{n-1}$

so $9x + 266$ is divisible by 7

as $266$ is divisible by $7$ so $9x$ is divisible by $7$ and $7$ is coprime to $9$ so $x$ is divisible by $7$

 

2025/004) prove that $\tan \, 1^{\circ}$ is irrational

 We know $\tan\,2A = \frac{2\tan\, A}{1+\tan ^2 A}\cdots(1)$

So if $\tan\, A $ is rational then $\tan 2A$ is rational 

Further $\tan(A+B) = \frac{\tan \, A + \tan\, B}{1-\tan\,A \tan \,B}\cdots(2)$

So if $\tan\, A$ and $\tan \, B$ are rational then $\tan(A+B)$ is rational

So if $\tan\, A$ and $\tan \,nA$ are rational then $\tan(n+1)A$ is rational

using Above 2 we get if $\tan\, A $ is rational then $\tan nA$ is rational for all $n \in Z$

if $\tan \, 1^{\circ}$ is rational then $\tan \, 30^{\circ}$ or $\frac{1}{\sqrt{3}}$ is rational which is not 

Hence $\tan \, 1^{\circ}$ is irrational 

Proved 

 

 

2025/003) If $\frac{1}{a + b}$, $\frac{1}{b + c}$ and $\frac{1}{c + a}$ are in A.P., prove that $b^2, a^2$ and $c^2$ are in A.P.

We have from the properties of AM

$\frac{1}{b + c} - \frac{1}{a + b} =   \frac{1}{c + a} - \frac{1}{b + c}$

Or $\frac{a-c}{(b+c)(a+b)} = \frac{b-a}{((a+c)(b+c)}$

Or $\frac{a-c}{a+b }= \frac{b-a}{a+c}$

Or $(a-c)(a+c) = (a+b)(b-a)$

Or $a^2-c^2 = b^2-a^2$ 

Or $b^2+c^2= 2a^2$   hence $b^2, a^2$ and $c^2$ are in A.P.

2025/002) What is the general solution of the linear diophantine equation $63x-23y=-7$

 Because $GCD(63,23) =1 $ so this has a solution. 

Now let us find a particular solution a and b such that $63a + 23b= 1$

For this we shall use extended Euclidean algorithm.

We have

$63= 2 * 23 + 17\cdots(1)$

$23= 17  + 6\cdots(2)$

$17= 2 * 6 + 5\cdots(3)$

$6= 5 + 1\cdots(4)$

as we have got 1 as the last remainder we are done,.

Now let us work back wards to see

$1= 6- 5$

$=6- (17 - 2 *6)$ from (3)

$= 3 * 6 - 17$

$= 3 * (23-17)  - 17$ from (2)

$= 3 * 23- 4 * 17$

$= 3 * 23 - 4 * (63-2 * 23)$ from (1)

$=11 * 23 - 4 * 63$

So we have $ 1= 11 * 23 - 4 * 63$

Hence $ -7 = - 77 * 23 + 28 * 63$

The above solution is correct however we can reduce the absolute value of coefficients of 23 and 63 by adding 63 to 1st one and subtracting 23 from 2nd one to get

$-7 = 5 * 63- 14 * 23$

so $(x,y) = (5,14)$ is a solution and general solution is $5 + 23t,14+63t)$ as $23 * 63 - 63 * 23 = 0$ 

2025/001) If $GCD(a,35) = 1$ then show that $a^{12} \equiv 1 \pmod {35}$

 We have $35 = 7 * 5$ so

$GCD(a,5) = 1$ and $GCD(a,7) = 1$

Because GCD(a,5) is 1 so as per Fermats Little Thorem $a^{4} \equiv 1 \pmod {5}$ or 

$a^{12} \equiv 1 \pmod {5}\cdots(1) $

Because GCD(a,7) is 1 so as per  Fermats Little Thorem $a^{6} \equiv 1 \pmod {7}$ or

 $a^{12} \equiv 1 \pmod {7}\cdots(2)$

From (1) and 2

 $a^{12} \equiv 1 \pmod {35}$