practice problems pending

 1) Find the largest natural number 'm' such that:

 (m^2) is the difference of two consecutive perfect cubes.
AND
 (2m + 79) is a perfect square.

S1)
m^2 = (n+1)^3 - n^3 = 3n^2 + 3n + 1

Since m,n are both natural numbers, RHS will grow much faster than LHS and at some point it will irreversibly exceed it.

m^2 - 1 = 3n(n+1)
So RHS is a multiple of 6
=> m^2 = 1 mod 6
m^2 = 25,49 satisfy this but
24 = 3n(n+1) => 8 = n(n+1) no go.
48 = 3n(n+1) => 16 = n(n+1) no go.

what next?
So let's try some other approach.
Once you complete the square in 'n', you will get:
4m^2 = 3(2n+1)^2 - 1
(2m-1)(2m+1) = 3(2n+1)^2

LHS has 2 consecutive odd numbers which are co prime. Why?
gcd(2m-1,2m+1) = gcd(2,2m-1) = 1

So
RHS should split cleanly into 2 factors which don't share any prime factors.

Case 1:
2m-1 = a^2
2m + 1 = 3b^2

Case 2:
2m + 1 = a^2
2m - 1 = 3b^2

Case 2:
2 = a^2 - 3b^2
=> a^2 = 3b^2 + 2
=> a^2 mod 3 = 2 which is not possible since it can be only 0,1

Case 1:
2m = a^2 + 1
Given: 2m + 79 = k^2 = a^2 + 80
(k-a)(k+a) = 80
2m = a^2 + 1 => 'a' is odd => 'k' is odd.
=> k-a and k+a are even.

Case 1.1
k+a = 40
k-a = 2
k = 21, a = 19
=> 2m = 19^2 + 1 = 362 => m = 181
2m + 79 = 362 + 79 = 441 is a perfect square.
m^2 - 1 = 3n(n+1) = 181^2 - 1 = 182 * 180 => n(n+1) = 182*60 = 2*91*60 = 2*7*13*2*2*3*5 = 2^3.3.5.7.13
n(n+1) = 2^3.3.5.7.13 = 8*13.15.7 = 104*105
So it works.

Case 1.2
k+a = 20
k-a = 4
k = 12, a = 8
k,a should be odd. Reject.

Case 1.3
k+a = 10
k-a = 8
k = 9, a = 1

Answer m = 181

2)

 a, b, c, d are integers.


M = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)

Show that 12 | M. (That is, prove that (12) divides (M).)

S2)
WLOG
a > b > c > d

Now, let's try to make it impossible to get factors of 2,3 from these terms.
So a-b, a-c, a-d are not divisible by 2 and 3 to start with.
=> Each of them is 1 mod 2 and 1/2 mod 3.
=> Parity of a(odd/even) is exactly opposite to that of b,c,d.

So b,c,d have same parity(odd/even).
=> (b-c)(b-d) is divisible by 4.

Let's see if it always holds by doing different parity groupings.
(a) (b,c,d) done.
(a,b,c,d) If all have same parity. Done.
(a,b) (c,d) Groups of 2 have same parity => (a-b) and (c-d) = 0 mod 2.

So divisibility by 4 always holds.

Similarly let's do it for 3.
By pigeonhole principle at least 2 of a,b,c,d will have same remainder by 3.
So will get a factor of 3 by those 2, e.g. if a,b have same remainder then a-b = 0 mod 3.


3)
Show that product of n consecutive natural numbers is divisible by n!
S3)
n consecutive natural numbers mean one of them will be divisible by each of 1,2, ... n
Because each nth number is div by 'n' and so on..


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