Below, we present a problem from the 1/7 Algebra and Number Theory set, along with 3 student submitted solution. You may vote up for the solutions that you think should be featured, and should vote down for those solutions that you think are wrong (voting is anonymous!). Also, feel free to make remarks about these solutions, especially since threading of comments has been introduced :).

Solving a quadratic sum If \(a+b+c = 0\) and \(a^2 + b^2 + c^2 = 22 \), what is \(a^4 + b^4 + c^4\)?

You may try the problem by clicking on the above link.

All solutions may have LaTeX edits to make the math appear properly. The exposition is presented as is, and have not been edited.

\[ \mbox{Remarks from Calvin} \]

Solution A - As Qi Huan mentions, this merely presents a specific case, and presumes that the numerical answer is uniquely determined due to the answer format used. This does not constitute a proof, unless you can further argue that \(a^4 + b^4 + c^4\) is a constant subject to the other conditions.

Solution B - Each step in this solution is necessary, and all these steps provide a complete solution. This solution is presented by Diego.

Solution C - Please check your that your algebra is correct before you submit a solution. If you expect me to read through your solution, you should have the courtesy to read through it once, and ensure that stupid mistakes are not made. All too often, proofs are wrong because of major mistakes with notation and algebra that could easily have been avoided.

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## Comments

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TopNewestSolution B - \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=0\)

\( 22+2(ab+bc+ac)=0 \Rightarrow ab+bc+ac=-11\)

\((ab+ac+bc)^2=a^2b^2+a^2c^2+b^2c^2+2(a^2bc+ab^2c+abc^2)\)

\( =a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c) \Rightarrow a^2b^2+a^2c^2+b^2c^2 =(-11)^2=121\)

\((a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)=484\)

\(a^4+b^4+c^4+242=484\)

\(a^4+b^4+c^4=242\)

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Who submitted the solution B?

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Solution B is correct. Solution C is incorrect, since a^4+b^4+c^4 is NOT equal to (a^2+b^2+c^2)^2-2ab-2bc-2ca. Solution A uses a special case,hence is not rigorous.

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Remarks about the solutions have been added to the discussion post.

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Solution A - Knowing that the cuaetic sum will be constant for all allowed points, we can play around with the values a bit. For example, by setting \(c=0\), we simplify our system of equations to: \[ 1. a + b = 0, \quad 2. a^2 + b^2 = 22\] Squaring the first equation and subtracting the second from it, we get \(2ab=-22\), or \(ab=-11\). Since \(a=-b\) by the first eauation, by substituting \(-b^2 = -11\), or \(b=+ \sqrt{11}\) or \(- \sqrt{11}\) .Since \(a\) and \(b\) are opposites, both have the same solution set but are different signs in each. Without loss of generality, assume \(a= \sqrt{11}\) and \(b=- \sqrt{11}\) , so the quartic sum is \(11^2 + 11^2 + 0 = 242\).

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Solution C - \(a=-b-c\)--->1 \(a^4+b^4+c^4=(a^2+b^2+c^2)^2-2ab-2bc-2ca =(22)^2-2(ab+bc+ca) =484-2(ab+bc+ca)\) sub 1 in the above eqn we get \(ab+bc+ca=121\) therefore \(484-242=242\).

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