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