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Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$

I tried adding and subtracting both equations, but didn't get anywhere. Would appreciated any ideas. Thanks!

user526427
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    I solved the two equations numerically, and the answers I got are $a\approx2.733$ and $b\approx-0.7311$. $a^4+b^4$ doesn't come out as anything special looking, it is $56.08$ at these values of $a$ and $b$. – Suzu Hirose Aug 24 '22 at 03:38
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    Show us your effort.Please include all the details that you have already done. By the way,See here – Sourav Ghosh Aug 24 '22 at 04:39

1 Answers1

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Adding both equation gives: $$a^4+b^4+8(a+b)=4(a^3+b^3)-8$$ i.e: $$(a^2-2a)^2-4(a^2-2a)+b^4-4b^3+8b+8=0$$

Consider this equation is a quadratic equation with variable $a^2-2a$

So the discriminant is:$(-2)^2-(b^4-4b^3+8b+8)=-b^4+4b^3-8b-4=-(b^2-2b-2)^2$

Then:$b^2-2b-2=0$, so $b=1-\sqrt{3}$ or $b=1+\sqrt{3}$ then you can easily find $a=1+\sqrt{3}$ and $b=1-\sqrt{3}$ is only solution which is satisfied. Then $a^4+b^4=56$

OnTheWay
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    It's not clear why the discriminant is zero. I think a better way is to get the expression as sum of two squares. The sum is zero only if each term is zero. You will get the same answer. – Andrei Aug 24 '22 at 03:44
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    I assume $a,b$ are real numbers, then for existing of $a,b$ then the quadratic equation must have at least one root, but $-(b^2-2b-2)^2\le0$ then $-(b^2-2b-2)^2=0$. You can use your method, i use mine. – OnTheWay Aug 24 '22 at 03:50
  • Indeed. That was the step that I've missed (even though I assume the same thing). +1 – Andrei Aug 24 '22 at 03:55
  • Exactly the same approach. See AoPS – Sourav Ghosh Aug 24 '22 at 04:40
  • In fact, your method is clearer, by splitting $8$ to $4+4$ and add them to both expressions to get $(a^2−2a-2)^2+(b^2−2b−2)^2=0$ which implies : $a^2−2a-2=0,b^2−2b−2=0$ – OnTheWay Aug 24 '22 at 04:48