2022/064) How to solve the Diophantine equation ab+a+b+82=z2, where (a,b,z) are positive integers & g.c.d(a,b)=1?

We have $ab + a + b + 82 = z^2$

or $ab + a + b + 1 + 81 = z^2$ we do so that we can facrot LHS

or $(a+1)(b+1) = z^2 - 81 = (z-9)(z+9)$ 

we can choose a +  1 = z - 9 and b+1 = z + 9 (this shall not give all solutions but some 

so a = z - 10 and b = z + 8

they have to be co-prime so both are odd say z = 2 m + 1

so a = 2m - 9 and b = 2m + 9

for these to be co-prime there should not a factor of 18 (that is the difference)

so 2m - 9 shall not have a factor 3 or m must not have a factor 3

so a = 2m - 9 and b = 3m + 9 m > 5 and of the forn $3k \pm 1$

 

2022/063) How can I find all positive integer values of N such that N(N−101) is a square of a positive integer

 Now let us take GCD(N,N-101)

GCD(N,N=101) = GCD(N,101)

There are 2 cases either N is multiple of 101 or not

If N is multiple of 101 say N= 101m

So N-101 = 101(m-1) so $N(N-101) = 101^2m(m-1)$ which is not a perfect square

So N is not a multiple of 101 so GCD(N,N-101) = 1 so N and N-101 both are perfect squares

Say $N= x^2$ and $N-101$ = y^2

$x^2-y^2 = 101$ or $(x+y)(x-y) = 101$ 

So x + y = 101 and x-y =1 solving these we get x = 51 and y = 50

So $N = x^2 = 51^ = 2601$

2022/062) Prove that every number in the sequence 16,1156,111556,⋯ is a perfect square.

We have for $n^{th}$ number it is n 1s followed by n-1 5's and ending with 6

of it is n 1's follwed by n 5's and add 1

n '1s $\frac{1}{9}(10^n-1)$

n 5's is   $\frac{5}{9}(10^n-1)$

so the number is  $\frac{1}{9}(10^n-1) * 10^n + \frac{5}{9}(10^n-1) + 1$

$= \frac{1}{9}(10^{2n}-10^n +  5* (10^n-1) + 9)$

$= \frac{1}{9}(10^{2n} +  4* 10^n + 4)$

$=\frac{1}{9} (10^n + 2)^2$

$=(\frac{1}{3}(10^n+2))^2$

which is  peffect square 

2022/061) What are the values of n for which $2^4+2^7+2^n$ is a perfect square?

 We have $2^4+2^7 + 2^n = 2^n + 144 = m^2$ where m is positive 

or $2^n = m^2 - 144 = (m+12)(m-12)$

as 2 is a prime so both m+ 12 and m-12 powers of 2

now difference (m+12)- (m-12) = 24

so powers of must have a difference 24

as $2^5 = 32$ so $2^6-2^k \ge 2^6 - 2^5 > 32$ (for any k less of equals 5)

so we need to check for candidates 2,4,8,16,32 and get 8 and 32.

this gives m = 20 and putting m = 20 we get n= 8

2022/060) If $x^2 - 3x +1 = 0$ and $x^{16} - k x^8 + 1 = 0$ then find k

Let $y^2- my + 1 = 0\cdots(1)$

so $y^2 = my -1$

or squaring both sides $y^4 = m^2y^2 - 2my + 1 = m^2y^2 - 2(y^2) + 1= (m^2-2)y^2$

or $y^4 = (m^2-2)y^2 + 1\cdots(2)$

putting y =x and m = 3 we get

$x^2 - 3x + 1$

and from (2) we have $x^4 - 7x^2 +1 = 0\cdots(3)$ 

putting $y = x^2$ and m = 7 in (1) we get (3) and from(2)  we get 

$x^8 - 47x^4 +1 = 9\cdots(4)$

  putting $y = x^4$ and m = 47 in (1) we get (4) and from(2)  we get  

2022/059) If $4x=a^2+3$ where (a,x) are positive integers & x is not a multiple of 3, then It's only possible when x are certain positive integers of the form of 6y+1. Prove this?

 Because 4x is even so we have a is odd 

So let a = 2n + 1

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

or $x= n^2 + n + 1 = n(n+1) + 1$

now n can be 0/1/2 mod 3

if n is one of the form $(3k, 3k+1, 3k-1$

If n is of the form 3k then $x-1 = 3k(3k+1)$ divisible by 6 so x of the orm 6m + 1

if n is of the form 3k + 1 then $x-1 = (3k+ 1)(3k+2) = 9k^2 + 9k^2 + 2 = 9k(k+1) + 2$ remainder 2 when devided by 6 

so x = 6m + 3 but is divisible by 3 so not admissible

if n is 3k -1 then 

$x-1 = (3k-1)(3k)$ of the form 6m or x+ 1of the form 6m + 1

So it is of the form 6m + 1(or 6y + 1)  and hence proved