Math Is Fun Forum

atreyyu

Member

Registered: 2008-08-01

Posts: 5

Solutions of a set + determine integers

Hey guys! I have 2 algerbra problems and I have no idea even where to begin . So could you please solve these problems and *explain* the solution to me?

1) Determine the number of solutions of the following set, depending on the parameter a.

2) Determine all positive integers

such that

is a prime number, and

is divisible by 3.

Thanks!

Last edited by atreyyu (2008-10-16 07:28:16)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Solutions of a set + determine integers

1.) if a=-1, then the graph looks like the letter V just below the origin.
Connect the points (-1,0) to (0,-1) and continue to (1,0).  That is the V shape.

And If   -1<a<+1, then there are 2 single points above the x-axis that
satisfy both equations.  (These double points always fall on the ^ shaped (hat-shaped)
graph formed by connecting (-1,0) to (0,1) and continue to (1,0).

And If a=+1, there is only one point, (0,1)

Last edited by John E. Franklin (2008-08-02 00:38:44)

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igloo myrtilles fourmis

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Solutions of a set + determine integers

(1) The graph of

is a square with vertices (÷1,0), (0,÷1). The graph of

is V-shaped symmetrical about the y-axis. In order for the equations to have solutions, we must have

. (There are no solutions for a outside this range.)

If

, there is exactly 1 solution, namely

.

If

, there are exactly two solutions, which are the intersection of

with the top half of

. The top half of the former is given by

. Equating,

, so

. Hence the solutions are

If

, there are infinitely many solutions: Every point on the lower half of the curve

is a solution. So the solution set is

.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Solutions of a set + determine integers

(2)

So you want to find positive integers a, b such that (i)

or (ii)

.

(i) is easy:

.

For (ii) assume that

; then if

is prime it’s not a multiple of 3.

Suppose a = 3k for some integer k. Then

and in order for this to be divisible by 3, b would have to be divisible by 3. Similar, if b were divisible by 3, so would a. So neither a nor b must be divisible by 3.

So we must have

for some

.

If

, then

, and if you simplify that, you’ll find that it’s congruent to 1 modulo 3. So that’s not divisible by 3. And

would similarly lead to

. So these cases are out.

If

, then a+b would be divisible by 3, and we don’t want that either.

Hence case (ii) is impossible. The only solutions are

.