Another positive integer puzzle (Posted on 2006-11-29)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
We can assume that gcd(a,b,c,d) =1. Otherwise, the equation can be simplified whenever gcd(a,b,c,d) is greater than 1. Clearly, d is divisible by 6. So, substituting d = 6m, we obtain: 6a2+ 3b2 + c2 = 30m2 Replacing c=3n in the above equation, we obtain: 2a2+ b2 + 3n2 = 10m2......(#) Now, we know that: 2a2 Mod 8 = 0, 2 2b2 Mod 8 = 0, 2 3n2 Mod 8 = 0, 3, 4 10m2 Mod 8 = 0, 2 By trying for each of the above quadruplet combinations, we observe that (#) is satisfied iff: (a2 Mod 8, b2 Mod 8, n2 Mod 8, m2 Mod 8) = (0,0,0,0); (2,2,0,0); (2,0,0,2); (0,2,0,2); (2,2, 4,0) Accordingly, each of a, b, n and m are even. This contradicts our original assumption that gcd(a,b,c,d) = 1 Consequently, no positive integer solution exists for the given problem. **************************************** PROBLEM SOURCE: Mathematical Olympiad Treasures by Titu Andreescu and Bogdan Enescu.

	Subject	Author	Date
	A start, perhaps?	tomarken	2006-11-29 21:57:53
	D2?	Richard	2006-11-29 21:32:47