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To cut a long story short:
Let the 3 numbers to be squared be a, b, c. By basic algebra:
I a^2+d=b^2
II a^2+2d=c^2
From I
III a^2+2(b^2-a^2)=c^2
IV 2b^2=c^2+a^2
This will always be true when a=b=c, but assume a and b are different.
From studying small examples in Excel, it seems likely that b is always a sum of squares, and that a= 2xy+(y^2-x^2), b=(x^2+y^2), c=2xy+(x^2-y^2).
Since d=c^2-b^2=b^2-a^2, we can test this:
(2xy+(x^2-y^2))^2-(x^2+y^2)^2 = 4(x^3y-xy^3) = 4xy(x-y)(x+y)=d
(x^2+y^2)^2-(2xy+(y^2-x^2))^2 = 4(x^3y-xy^3) = 4xy(x-y)(x+y)=d
demonstrating that this assumption is correct.
It is also obvious that 24 divides d (this was the subject of an earlier Perplexus which led to the current puzzle.)
Now we have b^2-a^2=24k, c^2-b^2=24k, and we want:
(c/x)^2-(b/x)^2=(b/x)^2-(a/x)^2=P, for prime P.
Rearranging, c^2-b^2=b^2-a^2=Px^2=24d. Since 24 is not a square, we need to start by multiplying by 6 to obtain 144 which is square.
If so, d = 4xy(x-y)(x+y)=144a^2P
Dividing through: xy^3-x^3y = (6a)^2P.
This should be sufficient to obtain the small solutions for 7 and 41 after a brief Excel search.
Setting x,y, square in xy(x-y)(x+y) , with (x+y) also square and (x-y) prime produces:
(414662879/4826640)^2-(407893921/4826640)^2 = 239
(407893921/4826640)^2-(401010721/4826640)^2 = 239
without undue effort.
Setting x,y, square in xy(x-y)(x+y) , with (x-y) also square and (x+y) a prime or square multiple of a prime leads quickly to several solutions e.g.
(56497/1560)^2-(49297/1560)^2 = 313
(49297/1560)^2-(40847/1560)^2 = 313
also found in Excel.
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