Fifthing the quart (Posted on 2013-04-28)

	Submitted by broll
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The problem is d2, so no need to overthink it! Rearrange the expression as a^4+d^4=b^5+c^5. A 4th power is also a square, so we have x^2+y^2=b^5+c^5, for some {x,y}. Since x and y are different, let y=(x+d). Now 2(x^2+(x+d)^2 = 2d^2+4dx+4x^2, which is 4x^2+4dx+2d^2. We can write this as d^2, plus 4x^2+4dx+d^2, or (2x+d)^2; twice a sum of squares is always a sum of squares. Since this is true of twice the 4th powers, it must be true of twice the 5th powers also.

	Subject	Author	Date
	re: Some (all?)	xdog	2013-05-01 19:24:27
	Some (all?)	Jer	2013-05-01 15:00:18