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Another positive integer puzzle (Posted on 20061129) 

Given that a, b, c, and d are positive integers, determine analytically if it is possible that
6(6a²+ 3b² + c²) = 5d².

Submitted by K Sengupta

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Solution:

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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:
6a^{2}+ 3b^{2} + c^{2} = 30m^{2}
Replacing c=3n in the above equation, we obtain:
2a^{2}+ b^{2} + 3n^{2} = 10m^{2}......(#)
Now, we know that:
2a^{2} Mod 8 = 0, 2
2b^{2} Mod 8 = 0, 2
3n^{2} Mod 8 = 0, 3, 4
10m^{2} Mod 8 = 0, 2
By trying for each of the above quadruplet combinations, we observe that (#) is satisfied iff:
(a^{2} Mod 8, b^{2} Mod 8, n^{2} Mod 8, m^{2} 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.
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PROBLEM SOURCE: Mathematical Olympiad Treasures by Titu Andreescu and Bogdan Enescu.

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