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Given quadratic equation $x^2+px+q+1=0$ with two distinct roots $x_1$ and $x_2$.

If $p$ and $p^2+q^2$ is prime numbers, what is the largest possible value from $x_1^{2014}+x_2^{2014}$?

My attempt: First of all, I think I need to find the formula for the simplest one like the sum and the product of its roots.

By using Vieta formula, I can get:

$$x_1+x_2= -p$$

$$x_1\times x_2= q+1$$

Then, I think I need to find another form like quadratic and cubic to help me to find the correct pattern for the desired question.

By using the algebra identity, I'll be able to find out that:

$$x_1^2+x_2^2=(-p)^2-2(q+1)$$

And $$x_1^3+x_2^3=(-p)^3-3(-p)(q+1)$$

Since the question is about 2014th degree, I think by using two identities is enough to find the pattern.

By setting the question, I got:

$$(x_1^{1007})^2+(x_2^{1007})^2= (x_1^{1007}+x_2^{1007})^2-2(x_1\cdot x_2)^{1007}$$

My question:

How can I get the value of $(x_1^{1007}+x_2^{1007})$? And also what is the useful for the info such that $p$ and $p^2+q^2$ are prime?

Thanks

Edit: p and q are also integers. The nature of its roots is whole number.

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  • $\begingroup$ I tried to compute $x_1^{1007}+x_2^{1007}$, but the result is ugly (very long). So, I think you should have a new approach. $\endgroup$ – GAVD Sep 8 '15 at 16:37
  • $\begingroup$ Is anything known about $q$? For example, is it an integer? $\endgroup$ – André Nicolas Sep 8 '15 at 16:43
  • $\begingroup$ Yes, p and p²+q² is prime numbers, so I think p and q must be an integer. Thanks $\endgroup$ – akusaja Sep 8 '15 at 16:48
  • $\begingroup$ $p^2+q^2$ being prime doesn't imply that $q$ is an integer. $\endgroup$ – joriki Sep 8 '15 at 17:16
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    $\begingroup$ I think you need to clarify if p, q are both integers, also about the nature of the 2 roots (are they complex numbers???) $\endgroup$ – user261263 Sep 8 '15 at 20:14
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Suppose p is odd. From $p^2 + q^2$ prime we have $p^2 + q^2=2$ or $p \not = q \ \text{mod} \ 2$. Because p prime then $|p| > 1$ so we cannot have $p^2 + q^2=2$. Therefore $p \not = q \ \text{mod} \ 2$ and, because p is odd, q must be even and $q + 1$ must be odd. Now from: $$ x_1^2+px_1+q+1=0 $$ we have $x_1(x_1+p)$ is odd so $x_1$ is odd and $x_1+p$ is also odd. Therefore the difference $x_1 + p - x_1$ must be even, and this is a contradiction with initial supposition that p is odd.

Conclusion: $p=2$ or $p=-2$. This case, the discriminant is $4 - 4(q+1) = -4q$, therefore $q \lt 0$.

Because $x_1, x_2$ are whole numbers, we have $x_1x_2 \ge 0$ so $q+1 \ge 0$.

It follows that $-1 \le q \lt 0$ so $q=-1$. The solution: $x_1=0, x_2 = 2$ gives $x_1^{2014} + x_2^{2014} = 2^{2014}$

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  • $\begingroup$ why does $x_1$ have to be an integer? $\endgroup$ – jschnei Sep 8 '15 at 20:07
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Notice

The question has been updated, the two roots $x_1, x_2$ are now required to be whole numbers. This answer no longer address this question properly. Since this answer still contains a small amount of interesting info, I will leave it here.


I suspect the sum is unbounded. Let $N_k$ be the $k^{th}$ Pythagorean prime, i.e prime that can be written as sum of two squares.

$$N_k = P_k^2 + Q_k^2\quad\text{ with }\quad P_k \ge Q_k$$

If one look them up on OEIS:

You will find a lot of $P_k$ is prime. Among the first $10000$ Pythagorean primes, $1883$ of the $P_k$ are prime. For example, we have $N_{9998} = 225149 = 443^2 + 170^2$ and $P_{9998} = 443$ and $N_k$ are both primes.

If we use $(P_k,Q_k)$ as $(p,q)$, since $P_k \ge Q_k$, one can verify aside from the first three $k$, the corresponding equation $x^2 + px + q+1$ will have two distinct real roots. As a consequence, among the first $10000$ Pythagorean primes, $1881$ of them generate a pair of $(p,q)$ which matches your requirement.

Let $x_{1,k}$, $x_{2,k}$ be the two distinct real root. WOLOG, we will assume $|x_{1,k}| \ge |x_{2,k}|$. Since $x_{1,k} + x_{2,k} = -P_k$, this leads to

$$x_{1,k}^{2014} + x_{2,k}^{2014} \ge x_{1,k}^{2014} \ge \left(\frac{P_k}{2}\right)^{2014}$$

If the list of primes in $P_k$ doesn't terminate, then $P_k$ will diverge. This mean the sum at hand will be unbounded....

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Let me try. One has $$x_1+x+2 = -p;\ \ x_1x_2 = q+1.$$

$$x_1^2+x_2^2 = p^2 - 2(q+1).$$

$$x_1^3+x_2^3 = -p^3+3p(q+1)$$

$$x_1^4+x_2^4 = p^4-4p^2(q+1) + 2(q+1)^2$$

$$x_1^{7} + x_2^{7} = (x_1^{3} + x_2^{3})(x_1^{4}+x_2^{4}) - x_1^{3}x_2^{3}(x_1+x_2) = [-p^3+3p(q+1)][p^4-4p^2(q+1) + 2(q+1)^2]+p(q+1)^3.$$

$$x_1^{8} + x_2^{8} = (x_1^{4} + x_2^{4})^2-2x_1^{4}x_2^{4} = [p^4-4p^2(q+1) + 2(q+1)^2]^2 - 2(q+1)^4$$

One can continue as follows.

$$x_1^{1007} + x_2^{1007} = (x_1^{503} + x_2^{503})(x_1^{504}+x_2^{504}) - x_1^{503}x_2^{503}(x_1+x_2).$$

$$x_1^{503} + x_2^{503} = (x_1^{251}+x_2^{251})(x_1^{252} + x_2^{252}) - x_1^{251}x_2^{251}(x_1+x_2).$$

$$x_1^{504}+x_2^{504} = (x_1^{252}+x_2^{252})^2 - 2x_1^{252}x_2^{252}.$$

$$x_1^{251} + x_2^{251} = (x_1^{125} + x_2^{125})(x_1^{126}+x_2^{126}) - x_1^{125}x_2^{125}(x_1+x_2).$$

$$x_1^{252} + x_2^{252} = (x_1^{126} + x_2^{126})^2-2x_1^{126}x_2^{126}$$

$$x_1^{125} + x_2^{125} = (x_1^{62} + x_2^{62})(x_1^{63}+x_2^{63}) - x_1^{62}x_2^{62}(x_1+x_2).$$

$$x_1^{126} + x_2^{126} = (x_1^{62} + x_2^{62})^2-2x_1^{62}x_2^{62}.$$

$$x_1^{63} + x_2^{63} = (x_1^{31} + x_2^{31})(x_1^{32}+x_2^{32}) - x_1^{31}x_2^{31}(x_1+x_2).$$

$$x_1^{62} + x_2^{62} = (x_1^{31} + x_2^{31})^2-2x_1^{31}x_2^{31}.$$

$$x_1^{31} + x_2^{31} = (x_1^{15} + x_2^{15})(x_1^{16}+x_2^{16}) - x_1^{15}x_2^{15}(x_1+x_2).$$

$$x_1^{32} + x_2^{32} = (x_1^{16} + x_2^{16})^2-2x_1^{16}x_2^{16}.$$

$$x_1^{15} + x_2^{15} = (x_1^{7} + x_2^{7})(x_1^{8}+x_2^{8}) - x_1^{7}x_2^{7}(x_1+x_2).$$

$$x_1^{16} + x_2^{16} = (x_1^{8} + x_2^{8})^2-2x_1^{8}x_2^{8}.$$

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