A problem that puts me in difficulty

[DorelXD]

Homework Statement

Let $a,b,c,d$ be real number that satisfy the property that $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$. Show that $a^n + b^n = c^n + d^n$ for any $n$, a natural number ( with $n > 0$ ) .

Homework Equations

$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1})$$

The Attempt at a Solution

I have two ideas but I couldn't succed to apply them properly:

1) First, I'm thinking at a mathematical induction, but it seems that in this case the proof via induction isn't so straightforward.

2) Second, I'm thinking of writing that : $a^n - c^n = d^n - b^n$ and, then then using the formula:

$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1})$$

But this doesn't seem very helpful either. The very first move I thought of making is writing that:

$a^n - c^n = d^n - b^n \to (a-c)(... )= (d-b)(...)$ and get rid of the first paranthesis. Unfortunately, I cannot do this because I don't know for sure if (a-c) and (d-b) are diffrent from 0.




Please, can sombeody guide me to the solution ? Are my ideas good ?

 

[ehild]

Homework Statement

Let $a,b,c,d$ be real number that satisfy the property that $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$. Show that $a^n + b^n = c^n + d^n$ for any $n$, a natural number ( with $n > 0$ ) .

Homework Equations

$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1})$$

The Attempt at a Solution

I have two ideas but I couldn't succed to apply them properly:

1) First, I'm thinking at a mathematical induction, but it seems that in this case the proof via induction isn't so straightforward.

2) Second, I'm thinking of writing that : $a^n - c^n = d^n - b^n$ and, then then using the formula:

$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + ... + x^1y^{n-2} + y^{n-1})$$

But this doesn't seem very helpful either. The very first move I thought of making is writing that:

$a^n - c^n = d^n - b^n \to (a-c)(... )= (d-b)(...)$ and get rid of the first paranthesis. Unfortunately, I cannot do this because I don't know for sure if (a-c) and (d-b) are diffrent from 0.

Please, can sombeody guide me to the solution ? Are my ideas good ?

The two properties $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ involve some simple relations between a,b,c,d. What are these? (Write out the equations as differences and factorize the second one. )

ehild

 

[DorelXD]

Well, I'd get that:

$$(a - c)(a+c) = (d-b)(d+b)$$

Now, if $a - c = 0$ then $d- b = 0$ , and $a = c$ and $b = d$ . In this case, the problem statement holds. So, let's asume that a -c and d - b are both different from 0 .
Then, I get that: $a + c = d + b$ .

Also, by squaring $a + b = c + d$ and using the problem statement I get that $ab = cd$

So, I found two new relations:
$$ab = cd$$
$$a + c = d + b$$

Is this correct ?

 

[DorelXD]

Wait a minute:

a + c = d + b
a + b = d + c

By summing the two relations we have that: 2a + b + c = 2d + b + c , so a = d .

If a = d then b = c and the statement is proved :) .

 

[ehild]

Well, I'd get that:

$$(a - c)(a+c) = (d-b)(d+b)$$

and you also know that a-c=d-b. **

Now, if $a - c = 0$ then $d- b = 0$ , and $a = c$ and $b = d$ . In this case, the problem statement holds.

well done! The other case is discussed in your next post.

ehild

 

[ehild]

Wait a minute:

a + c = d + b
a + b = d + c

By summing the two relations we have that: 2a + b + c = 2d + b + c , so a = d .

If a = d then b = c and the statement is proved :) .

Excellent! You have solved the problem :thumbs:

ehild

 

[DorelXD]

Thank you very much ! You really helped me. You gave me the right hint at the right time. All the best! Maybe we will see each other again around here! :) :)

 

[ehild]

See you soon

ehild

 

[Ray Vickson]

Wait a minute:

a + c = d + b
a + b = d + c

By summing the two relations we have that: 2a + b + c = 2d + b + c , so a = d .

If a = d then b = c and the statement is proved :) .

Another way: from $a+b = c+d$ and $a^2 + b^2 = c^2 + d^2$ we have
$$(a+b)^2 - a^2 - b^2 = (c+d)^2 - c^2 - d^2 \\<br /> \text{ or}\\<br /> 2a b = 2 cd \rightarrow ab = cd$$
Therefore,
$$a^2 + b^2 - 2 ab = c^2 + d^2 - 2cd \\<br /> \text{ or}\\<br /> (a-b)^2 = (c-d)^2 \rightarrow a-b = \pm(c-d)$$
Just consider separately the two cases $a+b = c+d, a-b = c-d$ and $a+b = c+d, a-b = d-c$.