Add To Cube, Get Square? (Posted on 2007-08-03)

	Submitted by K Sengupta
Rating: 4.0000 (1 votes)
Solution:	(Hide)
There do not exist any perfect cube that becomes a perfect square when 7 is added to it. EXPLANATION: By the conditions of the problem: n^2 = m^3 + 7, whenever n is an integer and m is a non negative integer. We will prove a stronger result, which states that there does not exist any integer pair (m, n) that satisfy n^2 = m^3 + 7. If possible, let m be even. Then, n^2 (Mod 4) = 3, which is a contradiction, since n^2 (Mod 4) = 0 or 1 accordingly as n is even or odd. Accordingly, m is odd, so that n^2(Mod4) = 0, 2 accordingly as m( Mod 4) = 1 or 3. But since there cannot exist any integer n satisfying n^2(Mod4) = 2, it now follows that n is even and m(Mod 4) =1 ......(i) Now, m^3 + 8 = (m + 2)(m^2 – 2m + 4), so that: n^2 + 1 = (m + 2)(m^2 – 2m + 4), with: (m+2)(Mod 4) = 3 (from (i)) If all the prime factors of m+2 were congruent to 1(Mod 4), then we would have (m+2)(Mod 4) = 1, which is a contradiction. Accordingly, we must have at least at least one prime factor(f, say) of m+2 which is congruent to 3(Mod 4). Thus, we have n^2 (Mod f) = -1, with f (Mod 4) = 3. However, by Quadratic Reciprocity, the equation n^2(Mod f) = -1 is solvable if and only if f (Mod 4) = 1. This leads to a contradiction. Hence the result is proved. Consequently there do not exist any perfect cube that becomes a perfect square when 7 is added to it.

	Subject	Author	Date
	No Subject	K Sengupta	2023-06-21 07:43:27
	seoexpert	Cecilia	2023-06-21 06:52:35
	No Subject	K Sengupta	2023-06-08 23:54:52
	Sebastian	Cecilia	2023-06-08 06:46:12
	Henry	Cecilia	2023-05-20 15:13:26
	Killian	Cecilia	2023-05-15 08:11:16
	second step	xdog	2007-08-03 19:27:45
	first step	xdog	2007-08-03 10:55:31
	re: thoughts	Charlie	2007-08-03 10:07:33
	thoughts	Charlie	2007-08-03 09:37:38