2025/017) Solve in positive integers $x^2-xy+y^2 = 13$

we have 

$x^2-xy+y^2 = 13$

we can complete square by addition of some thing from $y^2$ to $x^2-xy$

to get $(x^2-xy + \frac{y^2}{4}) +  \frac{3y^2}{4} = 13$

or $(x-\frac{y}{2})^2 +  \frac{3y^2}{4} = 13$

or multiplying by 4 we get $(2x-y)^2 + 3 y^2 = 52$

as it is sum of positive numbers we get $3y^2 <=52$ or $ y <=4$'

$y =1$ gives $(2x-y)^2 = 49$ or $2x-y=7$ as -7 shall give -ve x so x = 4, y= 1

$y=2$ gives $(2x-y)^2= 40$ not a square

$y=3$ gives $(2x-y)^2 = 25$ or $2x-y =  5$ giving x = 4 , y = 3 and $2x-y = -5$ gives -ve x

$y=4$ gives (2x-y)^2 = 4 $, $2x-y=2$ gives x = 3, y = 4

 $2x-y=-2$ gives x = 1, y = 4

So we have solutions (4,1),(4,3),(1,4),(1,3)  

2025/016) For a prime $p >4$, how do you prove that $3^p-2^p \equiv 1 \pmod {42}$

 We have $ 42 = 2 * 3 * 7$

We need to show that $3^p-2^p=1   \pmod {2}\cdots(1)$

$3^p-2^p=1   \pmod {3}\cdots(2)$

and  $3^p-2^p=1   \pmod {7}\cdots(3)$ 

As $3^p$ is odd for any p and and $2^p$ us even we get

$3^p-2^p=1   \pmod {2}$

this is we have proved (1) 

 now p is odd so we have

$3^p$ is divisible by 3.

or $3^p \equiv 1 \pmod 3\cdots(4)$  

now let us look as $2^p$ we have as p is odd so p = 2k+1

$2^p = 2^{2k+1} = 4^k * 2$

so $2^p \equiv 2 \pmod 3\cdots(5)$

From (4) and (5) we have

$3^p- 2^p \equiv 1 \pmod 3$

We have proved (2)

Now we need to check for mod 7

as p is greater that 4 and a prime p is of the form 6k+1 or 6k + 5

Let us take the 2 cases 6k+ 1 and 6k-+ 5

first 6k+ 1

as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem 

$3^6 \equiv 1 \pmod 7$

so $3^{6k+1} = (3^6)^k .3 = 3 \equiv 3 \pmod 7$

 Simlilarly 

$2^6 \equiv 1 \pmod 7$

so $2^{6k+1} = (2^6)^k .3 = 2 \equiv 2 \pmod 7$

Or $3^p-2^p \equiv 1 \pmod 7$ 

So we have proved (3) for p = 6k+ 1

 Next t 6k+ 4

as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem 

$3^6 \equiv 1 \pmod 7$

so $3^{6k+5} = (3^6)^k 243 = 243 \equiv 3 \pmod 7$

 Simlilarly 

$2^6 \equiv 1 \pmod 7$

so $2^{6k+5} = (2^6)^k .32 = 32 \equiv 2 \pmod 7$

Or $3^p-2^p \equiv 221 \pmod 7$

Or   $3^p-2^p \equiv 1 \pmod 7$

So we have proved (3) for p = 6k+ 5

 We have proved (3) for both cases

as (1) (2) (3) all are proved hence Proved  

  

 

 

2025/015) Prove $4^{2n}+10n \equiv 1 \pmod {25}$

 We shall prove if by binomial expansion

we have $4^{2n}$

$= (5-1)^{2n}$

 $ = \sum_{k=0}^{2n}{2n \choose k}5^{2n-2}(-1)^{k}$

$ = \sum_{k=0}^{2n-2}{2n \choose k}5^{2n-2}(-1)^k -  {2n \choose 2n-1}*5 + 1$ separating last 2 terms

1st sum each term is divisible by $5^2$ that is 25

so we are left with 

  $4^{2n} \equiv  - 10n +1  \pmod {25}$

or  

   $4^{2n}+10n \equiv 1 \pmod {25}$ 

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.