Science:Math Exam Resources/Courses/MATH437/December 2006/Question 04

MATH437 December 2006

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Question 04
For which of the following values of k does $\displaystyle x^{2}+y^{2}=k!$ have a solution with x, y integers: (a) $\displaystyle k=6$ (b) $\displaystyle k=12$ (c) $\displaystyle k=24$ (d) $\displaystyle k=48$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall Fermat's Theorem on the Sums of Two Squares: A number n is a sum of two squares if and only if all prime factors of the form $\displaystyle 4k+3$ have even exponent in the prime factorization of n.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (a) Since $\displaystyle 6!=2^{4}\cdot 3^{2}\cdot 5$, Fermat's theorem says that in case (a) we must have a solution. (The curiously minded person should note that as $1^{2}+2^{2}=5$ then multiplying both sides by $12^{2}$ gives $\displaystyle 12^{2}+24^{2}=6!$). (b) As $\displaystyle 11\parallel 12!$ and $\displaystyle 11\equiv 3\mod {4}$, this case has no solutions. (c) As $\displaystyle 23\parallel 24!$ and $\displaystyle 23\equiv 3\mod {4}$, this case has no solutions. (d) As $\displaystyle 47\parallel 48!$ and $\displaystyle 47\equiv 3\mod {4}$, this case has no solutions.

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