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A Quintic Problem (Posted on 2006-03-15) |
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Determine all possible integer solutions, whether positive or negative, to this equation:
5y4 + 2560y2 = x5 - 65536
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Submitted by K Sengupta
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Rating: 2.3333 (3 votes)
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Solution:
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(Hide)
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x=16 and y = +/- 16 are the only possible integer solutions to the equation under reference.
EXPLANATION:
From the given equation, we obtain:
5*(y^4) + 2560*(y^2) + 65536 = x^5
or, 32(5*(y^4) + 2560*(y^2) + 65536) = P^5, where P=2x
or,(y+16)^5 = (y-16)^5 + P^5 .............(#)
Now, we know that in accordance with Fermat's Last Theorem, the Diophantine equation :
a^n + b^n = c^n
where a, b, c and n are integers, has no nonzero solutions for n> 2.
(Reference:http://mathworld.wolfram.com/FermatsLastTheorem.html).
Hence, for integers ( positive, negative or zero) a, b and c; the equation a^5 + b^5 = c^5 is satisfied only in the following cases:
Case (i) a=0 giving b=c, Case (ii) b=0 giving a=c Case (iii) c=0 giving a=-b Case (iv) a=b=c=0
In terms of equation (#), we can substitute, without any loss of generality:
a = y-16, b= P and c = y+16.
Now, in terms of case (i), y-16=0 and 2x = y + 16, so that y = 16 giving 2x = 32, so that x = 16. Accordingly, (x, y) = (16,16).
In terms of case (ii), P =0 , or 2x = 0, giving x = 0, which is not feasible as x must be a non-zero integer ( whether positive or negative ), by conditions of the problem. This is a contradiction.
In terms of case (iii), y+16 =0 and (y-16) = - P giving, y = -16 and (y - 16) = - 2x, so that: y = - 16 and x = -32/ -2 = 16.
In terms of case (iv), P =0 giving x=0, which is not feasible for reasons explained earlier.
Combining all the four cases we obtain x = 16 and y = +/- 16, which constitute the only possible solutions to the equation under reference.
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Note: This problem is inspired by a previous puzzle submitted by Nick Hobson.
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