## Wednesday, December 30, 2009

### Number of solutions

Here I start my blog with a very interesting problem.

#Problem 1

Find the value of x and y. x^3+y^3=7163.[Where x and y are positive integers]

Solution scheme and approach 1

First of all one number should be even another is odd. Let, x=(2n+1) and y=2m
So,
x^3+y^3=7163
=>(2n+1)^3+(2m)^3=7163.....(i)
=>8n^3+12n^2+6n+1+(2m)^3=7163
=>2n(4n^2+6n+3)+8m^3=7162
=>n(4n^2+6n+3)+4m^3=3581

As RHS is odd ans 4m^3 is even therefore n(4n^2+6n+3) must be odd =>n is odd=>n=2a+1

From eqn (i) we can write
(4a+3)^3+(2m)^3=7163
=>64a^3+144a^2+108a+27+8m^3=7163
=>64a^3+144a^2+108a+8m^3=7136
=>a(16a^2+36a+27)+2m^3=1784

As 2m^3 is even therefore a(16a^2+36a+27) should be even=>a is even

Now by trial and error we have to move forward.
If a=2
then a(16a^2+36a+27)=326 =>m^3=729 =>m=9

So (a,m)::(2,9) =>(x,y)::(4*2+3,2*9)=>(x,y)::(11,18)

Solution scheme and approach 2

x^3+y^3=(x+y)(x^2-xy+y^2)=7163=13*19*29

As x and y are positive integers therefore x+y=1,x+y=7129 should be rejected. and we can also write that x^2+y^2-xy>x+y

3 cases are possible
(i)If x+y=13
then, x^2-xy+y^2=(x+y)^2-3xy=169-3xy=19*29=>3xy=-382 (impossible)

(ii)If x+y=19
then (x+y)^2-3xy=13*29=>361-3xy=13*29 =>3xy=-16(impossible)

(iii)If x+y=29
the (x+y)^2-3xy=29^2-3xy=13*19=>3xy=841-247=>3xy=594=>xy=198=11*18

(x,y)::(11,18)

Hope you have enjoyed the problem :)

Problem for practice:

If ${a^4} + {b^4}$ =4721. Then find the sum of all values of (a+b).
(a)9, (b)11, (c)13, (d)15, (e)None of the foregoing [Source of this question is totalgadha.com/rmo]

1. keep up good work man.

2. for the sum a^4+b^4=4721..we can do by even odd concepts.But can be done littl diff way.

=>(a^2)^2+(b^2)^2=4721 i.e sum of two sq end with 1.

sq numbers end with 1,4,5,6,9,0.
so here we can get combination 0,1 and 5,6 for the sum end with 1.
if we take 0,1;its a problem coz for that a^2 or b^2 has to end with 0.so minimum value has to be 100.so not possible as 100^2>4721.
so only option left with 5,6
now if a^4 ends with 5 means a also ends with 5.hence the value a can take may be 5,15,25...
but only option is 5 as rest exceed.
we can similarly apply same logic to get b^4 ends with 6..and b ends with 4,6 or 8..n hence applying similar logic we get 8 is the only option.(it should be done to ensure that there is no other pair).
but better in a quick note 4721-5^4=4096=8^4.
so (a,b)=(5,8) or (8,5)

if we consider the ordered pairs its 26,but non ordered pairs,its 13.