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