Elementary problems that still baffle me

[Gib Z]

Well There are numerous problems, I'll post up a few at a time. This is part of a larger assignment and I am so ashamed to say that I don't have any idea on some of them.

1) Factor $x^8+2x^4y^4+9y^8$
and $a^4+b^4+c^2 - 2(a^2b^2 +a^2c+b^2c)$

I thought maybe it was some perfect square, the first one. I noticed the powers were quite coincidental so i let u=x^4 and v=y^4, but it seems there are no numbers that multiply to 9 and add to 2...I think that's irreducible. The second one I hoped I would notice its some expansion of something familiar, but I didnt get it..

Edit: I just tried Mathematicia and get $(x^4-2x^2y^2+3y^4)(x^4+2x^2y^2+3y^4) \mbox{and} (a^2-2ab+b^2-c)(a^2+2ab+b^2-c)$ But I want to know how to get them manually!

2) Prove $20^{22}-17^{22} +4^{33} -1$ is divisible is 174.

I think i might need some mod arithmetic or something here, but I haven't done that before. Also, I noticed the differences, 20-17, 4-1, are the same, 3, which 174 is divisible by. And perhaps even more useless, that 174 is a 17 and 4 put together, which are in there...I tried expanding 20^22-17^22 and 4^33 -1 by $x^n-1=(x-1)(1+x+x^2+x^3\cdots +x^{n-2}+x^{n-1})$ and the similar one for x^n-y^n, but no go.

Edit: Once again mathematicia says the actualy number is 40768477197265827835864143774, and when divided by 174 gives 234301593087734642734851401, but I want to know how to do it.

3) Prove that if p,q,r and s are odd integers, then $x^{10}+px^9-qx^7+rx^4-s=0$ has no integer roots.

Just no idea on that one, none at all.

4) For which real values of b do the equations $x^3+bx^2+2bx-1=0 \mbox{and} x^2+(b-1)x+b=0$ have a common root?

I tried using the quadratic formula on the quadratic, and the discriminant was b^2-6b+1, which doesn't seem to help me in any way.

5) Prove $a^4+b^4+c^4 >= a^2bc+b^2ac+c^2ab$ for all real a,b,c.

Now usually when I encounter these type of questions It always happens to work that I minus the RHS from both sides, and the LHS can be factorized into some perfect square hence more or equal to zero. But I can't factorize it this time!

I know that if I get desperate I would use some multivariable calculus and find the global extrema of the function, But I would like to know how to do this with precalc methods thanks..6) Prove that if real numbers a, b and c satisfy $a+b+c > 0, ab+ac+bc>0, abc > 0$ then each a, b and c are positive.

I can't believe I can't do these...Sorry for the hassle guys, thanks!

 

[Flux = Rad]

OK, so I don't know much about rigorous mathematical proofs but I can kinda intuitively prove 6).

$$abc > 0$$ says that
-all 3 terms are positive
or
-one is positive and two are negative.

Let's suppose 2 of them are negative.
$$a + b + c > 0$$ says that the positive one is greater than the |sum of the negative ones|.

If we now look at $$ab + ac + bc > 0$$, only one of the terms is positive. Furthermore it's the product of the two smaller numbers. So it clearly can't be true for our initial assumption of 1 positive, 2 negative.

So they must be all positive.

 

[Gib Z]

I don't get how you finished it... Let's just say we choose b and c to be the negative ones. then a(b+c) + bc > 0 ----> -a|b+c| > -bc ----> a|b+c| > bc

But that doesn't get me the answer! the negative ones arent "smaller" is magnitude, just negative. Thanks for the help so far.

 

[Flux = Rad]

When you've got $$-a|b+c| > -bc$$, if you want to remove the minus signs you have to flip the inequality sign. So you end up with $$a|b+c| < bc$$.

OK, now consider $$a + b + c > 0 \Rightarrow a > |b + c|$$.

We had $$a|b+c| < bc$$, so now that we know that $$|b + c| < a$$, we can replace $$a$$ with $$|b + c|$$ and our inequality will still have to be true.

This says $$|b + c|^2 < bc$$. Expanding out and taking all the terms to one side we get the following contradiction: $$b^2 + bc + c^2 < 0$$

 

[Gib Z]

Ok I get most of it, but the "Expanding out" bit confused me. Did you use the triangle inequality or something? If you did, i don't think it could work because |b+c| </= |b|+|c|, so that so the directions of the arrows don't match up if you get what i mean.

 

[Hurkyl]

3) Prove that if p,q,r and s are odd integers, then $x^{10}+px^9-qx^7+rx^4-s=0$ has no integer roots.

Just no idea on that one, none at all.

Well, the right hand side is even.

 

[nrqed]

Well There are numerous problems, I'll post up a few at a time. This is part of a larger assignment and I am so ashamed to say that I don't have any idea on some of them.

1) Factor $x^8+2x^4y^4+9y^8$
and $a^4+b^4+c^2 - 2(a^2b^2 +a^2c+b^2c)$

I thought maybe it was some perfect square, the first one. I noticed the powers were quite coincidental so i let u=x^4 and v=y^4, but it seems there are no numbers that multiply to 9 and add to 2...I think that's irreducible. The second one I hoped I would notice its some expansion of something familiar, but I didnt get it..
I can't believe I can't do these...Sorry for the hassle guys, thanks!

If we focus on x as the unknown, notice that the polynomial contains x^8, x^4 and an x independent piece. This shows that you may use the quadratic formula for the variable x^4. This trick woul dwork also if you had an x independent term, a x^3 and a x^6, etc. Notice that it works also for the second polynomial since it contains a^4, a^2 and a independent terms.

 

[VietDao29]

...
1) Factor $x^8+2x^4y^4+9y^8$
and $a^4+b^4+c^2 - 2(a^2b^2 +a^2c+b^2c)$

For the first one, you can try to complete the square, like this:
x⁸ + 2x⁴y⁴ + 9y⁸
= (x⁴ + 3y⁴)² - 4x⁴y⁴.
Hopefully, I think you can go from here, right? :)

The second one can be done in the same manner, notice that:
(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc.

When you see something to the power of an even number (i.e 2n), and something else to the power of half of that number (i.e, n), and some have the factor 2 in front of it. You should think about completing the square.

 

[NateTG]

Hmm... 174=2*3*29 so if you show that
$$20^{22}-17^{22}+4^{33}-1$$
is divisible by 2,3, and 29, then you're done.
Each of those is individually easy using modular arithmetic.

 

[Gib Z]

Thanks so much for the help so far guys.

Hurkyl -

If even, the 4 first terms of the polynomial must evalute to an even number as well. this implies s is even, but s is odd. Contradiction.

If odd, the the 4 first terms also evaluate to an even number, same contradiction. Thanks for that hint :D

nrged- I tried the quadratic formula but I also get a discriminant term that is not a perfect square and I can't simplify it, so it doesn't help me.

VietDao29- Completing the square is what I should have thought of first :) Thanks for that, Differences of 2 squares after that :) As for the expansion of the (a+b+c)^2, would it be better to just know that or should I have worked it out this time?

NateTG- I haven't done modular arithmetic before, But ill go on wikipedia and try it now. Thanks

 

[Hurkyl]

NateTG- I haven't done modular arithmetic before, But ill go on wikipedia and try it now. Thanks

Actually, you have. Even/odd is the simplest nontrivial example of modular arithmetic.

If even, the 4 first terms of the polynomial must evalute to an even number as well. this implies s is even, but s is odd. Contradiction.

If odd, the the 4 first terms also evaluate to an even number, same contradiction. Thanks for that hint :D

 

[Gib Z]

I know what it is now, But I can't apply it :'( So its like the remainder, but the powers on the numbers confuse me...

 

[NateTG]

I know what it is now, But I can't apply it :'( So its like the remainder, but the powers on the numbers confuse me...

Let's try it this way:
Let's say we want to know what the remainder is when we divide
$17^{22}$ by $29$
We know that
$$17^{22}=17^{16}*17^4*17^2$$
(The exponents on the RHS are powers of 2 which will be convenient in a moment.)
Now, it's easy to work out that
$$17*17=289=29*9+28$$
and then
$$17^4=\left(17^2\right)^2=(29*9+28)^2=(29*9^2)*29+(2*9*28)*29+(28*28)=(n)*29+(28*28)$$
We really don't care about $n$ since we're only looking for a remainder when dividing by 29, so that leaves
$$28*28=784=(27)29+1$$
So
$$17^4=(m)29+1$$
now,
$$17^8=\left(17^4\right)^2=\left((m)29+1\right)^2=(29m^2+2m)29+1=(o)29+1$$
and similarly
$$17^16=\left(17^8\right)^2=(p)29+1$$
substitute those in in the equation:
$$17^{22}=17^{16}*17^4*17^2$$
$$17^{22}=\left(29(p)+1\right)*\left(29(n)+1\right)*\left(29(9)+28\right)=\left(29(q)+28\right)$$

So, clearly the remainder of $17^{22}$ divided by $$29$$ will be $28$.

This may have looked like a lot of work, but really, it was more effort to write stuff out than it was to do the multiplications ($17*17,28*28,1*1 \rm{and} 1*1*28$), or find the remainders.

A similar process will get you the remainders for the other terms, which can then be added together.

 

[Gib Z]

Thanks tonnes for that guys, especially NateTG :)

I used that method to prove the remainders for the division by 29. For 20^22 I got 20 if i remember, and for 4^33 i got 9 i think.

For divisibly by 2, that can use simpler arguments as you know.

For divisibility by 3, I sort of noticed that the bases in question, 20, 17 and 4, and only 1 off multiples of 3. So basically I used the method NateTG posted, but it was somewhat quicker.

eg $$(3+1)^{33} = ...\mbox{some terms with a factor of 3} ... + 1^{33}$$

When divided by 3, it obviously gives a remainder of 1. :D Similarly for the others. Thanks for the help on that one!

So now its just problems 4 and 5 left.

 

[Gib Z]

I just worked out another even easier way to see the divisibility by 3. It by expansion of x^n - y^n = (x-y)(blah blah...) - because in both cases 20-17 and 4-1, its 3. :) Since I haven't learned mod arithmetic yet, maybe there was some other way I could have seen divisibility by 29? Just might be easier that's all.

For number 4, I thought maybe solve the cubic with Cardano's method, but I am not meant to know that in class either...

 

[Gib Z]

Ok guys I just found the easiest and expected way I was meant to work out number 2 (the divisibility by 174 one). Basically we show its divisible by 3 and 58. We do 3 via the expansion in the post just before this one, and we do 58 like this:

$$20^{22}-17^{22} + 4^{33} - 1 = 20^{22} + 4^{33} - (17^{22}+1) = 400^{11} + 64^{11} - (289^{11} + 1^{11})$$

Then use the expansion (valid for odd n):
$$x^n+y^n=(x+y)(x^{n-1} - yx^{n-2} + y^2x^{n-3}...+y^{n-1})$$

The first factor, the (x+y) becomes 464 for the first pair, and that's divisible by 58. The 2nd factor, (x+y) becomes 290, gives is also divisble :)

But now I have only 3 days to solve number 4, any ideas because I am fully stumped..

 

[povatix]

What about the others?

These are great problems, but what about the others(3-6)?,it would be nice to know how to solve them :tongue:

 

[learningphysics]

For part 4, assume both equations to be true. try to eliminate terms to get simpler equations (eliminate high order terms)... sort of like the way you'd solve a system of linear equations...

 

[Kummer]

$$a^4+b^4+c^4 \geq a^2b^2+b^2c^2+a^2c^2$$
But,
$$a^2b^2+b^2c^2+a^2c^2\geq a^2bc+ab^2c+abc^2$$

Q.E.D.

 

[povatix]

thanks for the help guys

 

[Gib Z]

Sorry this is quite an old thread. I've finished the problems on my own since I started the thread, no one was posting here anymore so i thought no one cared lol.

Yes Kummer is correct, same method as me. 2 applications of AM-GM inequality, or simply using the fact x^2 + y^2 > 2xy twice.Ok You want Solutions for 3-6.

Post 2 and 4 solve number 6. Post 10 has number 3. Kummer just showed 5.

Problem 4 took me the longest, and I re wrote the LHS as $$x^{3}+bx^{2}+2b-1=(x+1)(x^{2}+(b-1)x+b)-(2b-1)x+(b-1)$$

So, if $$x$$ is root of $$x^{3}+bx^{2}+2b-1=0$$ and $$x^{2}+(b-1)x+b=0$$, it is also root of $$(2b-1)x-(b-1)=0$$ and so $$b\neq\frac{1}{2}$$ and $$x=\frac{b-1}{2b-1}$$
If we import this value in $$x^{2}+(b-1)x+b=0$$, we have $$\frac{(b-1)^{2}}{(2b-1)^{2}}+\frac{(b-1)^{2}}{2b-1}+b=0$$
So $$(b-1)^{2}+(b-1)^{2}(2b-1)+b(2b-1)^{2}=0$$
So $$b(2(b-1)^{2}+(2b-1)^{2})=0$$
And so the only possibility would be $$b=0$$ and we check easily that with $$b=0$$, we have $$x=1$$ common root of the two polynomials.

 

[povatix]

Thanks so much for the help so far guys.

Hurkyl -

If even, the 4 first terms of the polynomial must evalute to an even number as well. this implies s is even, but s is odd. Contradiction.

If odd, the the 4 first terms also evaluate to an even number, same contradiction. Thanks for that hint :D

thanks for the super quick reply, but can u please disambiguate the above for me, it sounds so simple, yet i cannot draw anything out of it..

 

[Gib Z]

Ok well Do you know the following facts?

An even number minus or plus another even number is still even, and plus/minus another odd number is odd?

An odd number plus/minus an odd number is an even number, and plus/minus an even number is still odd?

An even number to an integer power is still even?

An odd number to an integer power is still odd?

Using those facts, and taking the "s" in the original question to the RHS: If x is equal to some EVEN number, then the LHS is EVEN, but we know the RHS, s, is odd. So x can't be even.

Same argument for x being Odd. The Odd's and Even's make the integers, so yea.

 

[povatix]

Thanks, it's all starting to rush into my brain. It's amazing how smart people can be.

 

[learningphysics]

Problem 4 took me the longest, and I re wrote the LHS as $$x^{3}+bx^{2}+2b-1=(x+1)(x^{2}+(b-1)x+b)-(2b-1)x+(b-1)$$

Your initial question gave $$x^{3}+bx^{2}+2bx-1$$ not $$x^{3}+bx^{2}+2b-1$$