What are the values of the six integers?
`x^2 = a^2 + 4b + 1``y^2 = b^2 + 4c + 1`
`z^2 = c^2 + 4a + 1`
a, b, c, x, y, z are all positive integers
a > b > c, x > y > z.
What are the values of the six integers? `b>c>0, y^2=b^2+4c+1 \\→ b^2+1<y^2<b^2+4b+1\\→b<y<b+2\\→y=b+1,b=2c`
同理，`x=a+1, a=2b`.
即 `a=4c, b=2c, x=4c+1, y=2c+1, z^2=c^2+16c+1` `z^2=c^2+16c+1 → (c+8+z)(c+8-z)=63`.
由`y>z`得 `(2c+1)^2>c^2+16c+1→c≥5, z≥11→c+8+z≥24`.
故`c+8+z= 63,c+8-z=1→ c=24, z=31`.
最后得` c=24, b=48, a=96, z=31, y=49, x=97`.
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