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¥Ñ scsnake ©ó ¬P´Á¥| ¤»¤ë 26, 2003 8:10 pm


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¥Ñ bye ©ó ¬P´Á¥| ¤»¤ë 26, 2003 8:25 pm


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¥Ñ scsnake ©ó ¬P´Á¥| ¤»¤ë 26, 2003 8:29 pm


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¥Ñ Raceleader ©ó ¬P´Á¤@ ¤»¤ë 30, 2003 10:17 am


Suppose there is a integral solution (x,y)=(y+k,y), where k is an integer.
(y+k)^2+5=y^3
y^3-y^2-2ky-k^2-5=0
y is 1 or -1 or 5 or -5 because the coefficients of the cubic equation are integral.
By putting the probable y into the equation, we solve k as non-integral which contradicts our condition k is integral.
So, there is no integral solution for x^2+5=y^3.

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