MHB Solve System Problem: x^4+y^4=8^2 & x-y=2

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The discussion revolves around solving the system of equations \(x^4+y^4=64\) and \(x-y=2\). Participants derive a quartic equation by substituting \(y=x-2\) into the first equation, leading to \(x^4 - 4x^3 + 12x^2 - 16x - 24 = 0\). The analysis reveals that this quartic has two real roots, which can be expressed as \(x = 1 \pm \sqrt{2\sqrt{10}-3}\). The factorization of the quartic confirms the existence of two real roots and two complex roots for \(y\). The conversation emphasizes the importance of verifying solutions against the original equations to avoid spurious results.
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Problem:
Solve the system $x^4+y^4=8^2$ and $x-y=2$.

Attempt:
$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

$(x+y)^4=(x^4+y^4)+4xy(x^2+y^2)+6x^2y^2$

$(x+y)^4=8^2+4xy((x-y)^2+2xy)+6x^2y^2$

$(x+y)^4=64+4xy((2)^2+2xy)+6x^2y^2$

$(x+y)^4=64+16xy+14x^2y^2$

Up to this point, I know that I need to form another equation (with $(x+y)^4$ the subject) that has only terms of xy...

$x-y=2$

$(x-y)^2=2^2$

$x^2+y^2-2xy=4$

$(x+y)^2-4xy=4$

$(x+y)^4=(4+4xy)^2=16+32xy+16x^2y^2$

$\therefore 64+16xy+14x^2y^2=16+32xy+16x^2y^2$

$x^2y^2+8xy-24=0$

And by using the quadratic formula to solve for xy, I get:

$\displaystyle xy=\frac{-8 \pm \sqrt {8^2-4(-24)(1)}}{2}=\frac{-8 \pm 4 \sqrt{10}}{2}=-4 \pm 2\sqrt{10}$

I then solve the values for y by multiplying each and every term of the equation $x-y=2$ by y, and obtained another quadratic equation in terms of y and from there, I managed to find all 4 pairs of answers accordingly.

Now, I was wondering if there are any other methods to tackle this problem effectively.

Thanks.
 
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I think you should double-check your answers against the original equation - you might have gained a few spurious solutions. This plot seems to indicate there should only be two solutions. Your solution method is very elegant, I'm not sure I know of a better.
 
Substitute $y=x-2$ in the equation $x^4+y^4 = 8^2$, you get a quartic equation for $x$, namely $x^4+(x-2)^4 = 64$, which simplifies to $x^4 - 4x^3 + 12x^2 - 16x - 24 = 0.$ Looking at Ackbach's plot of the functions, it appears that there are only two real roots for $x$, and it looks as though their sum is very close to $2$. That made me wonder whether the quartic might have a factorisation of the form $x^4 - 4x^3 + 12x^2 - 16x - 24 = (x^2-2x+a)(x^2-2x+b)$. Comparing coefficients, you can check that this is indeed the case, with $a = 4 - 2\sqrt{10}$ and $b = 4 + 2\sqrt{10}$. The first factor has two real roots $x = 1\pm\sqrt{2\sqrt{10}-3}$ (and the second factor has no real roots). So the solutions for $x$ are $\boxed{1\pm\sqrt{2\sqrt{10}-3}}$.

Edit. Since $y = x-2$, the factors $x^2-2x + 4\pm2\sqrt{10}$ above are just another way of writing the factors $xy + 4\pm2\sqrt{10}$ that anemone had already found. So all that was needed was to make that substitution $y = x-2$, and you get two quadratic equations for $x$, one with the two real roots and the other with the two complex roots.
 
Last edited:
Ackbach said:
I think you should double-check your answers against the original equation - you might have gained a few spurious solutions. This plot seems to indicate there should only be two solutions. Your solution method is very elegant, I'm not sure I know of a better.

Ops, I forgot to double-check the answers...thanks for reminding me and that's so nice of you to say that my solution method is elegant...

Thanks, Ackbach!:)

Opalg said:
Substitute $y=x-2$ in the equation $x^4+y^4 = 8^2$, you get a quartic equation for $x$, namely $x^4+(x-2)^4 = 64$, which simplifies to $x^4 - 4x^3 + 12x^2 - 16x - 24 = 0.$ Looking at Ackbach's plot of the functions, it appears that there are only two real roots for $x$, and it looks as though their sum is very close to $2$. That made me wonder whether the quartic might have a factorisation of the form $x^4 - 4x^3 + 12x^2 - 16x - 24 = (x^2-2x+a)(x^2-2x+b)$. Comparing coefficients, you can check that this is indeed the case, with $a = 4 - 2\sqrt{10}$ and $b = 4 + 2\sqrt{10}$. The first factor has two real roots $x = 1\pm\sqrt{2\sqrt{10}-3}$ (and the second factor has no real roots). So the solutions for $x$ are $\boxed{1\pm\sqrt{2\sqrt{10}-3}}$.

Edit. Since $y = x-2$, the factors $x^2-2x + 4\pm2\sqrt{10}$ above are just another way of writing the factors $xy + 4\pm2\sqrt{10}$ that anemone had already found. So all that was needed was to make that substitution $y = x-2$, and you get two quadratic equations for $x$, one with the two real roots and the other with the two complex roots.

Thanks, Opalg! Your approaches to the questions I post here at MHB are always surprisingly insightful to me.(Happy)
 
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