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解费尔马1的大勾股数方程:
(A^2)^2+(B^3)^2=(C^5)^2
即A^4+B^6=C^10
A=256*uv*(u^4-v^4)^(15k+10)*(u^4+v^4)^(15k+12)
B=32*(u^4-v^4)^(10k+7)*(u^4+v^4)^(10k+8)
C=8*(u^4-v^4)^(6k+4)*(u^4+v^4)^(6k+5)
其中,u、v为正整数,u>v,k为0或正整数。
验:令u=2,v=1,k=0;
A=256*uv*(u^4-v^4)^(15k+10)*(u^4+v^4)^(15k+12)=256*2*15^10*17^12
B=32*(u^4-v^4)^(10k+7)*(u^4+v^4)^(10k+8)=32*15^7*17^8
C=8*(u^4-v^4)^(6k+4)*(u^4+v^4)^(6k+5)=8*15^4*17^5
A^4=2^36*15^40*17^48=2^30*15^40*17^48*8^2
B^6=2^30*15^42*17^48=2^30*15^40*17^48*15^2
C^10=2^30*15^40*17^50=2^30*15^40*17^48*17^2
8,15,17是一组勾股数,
A^4+B^6=C^10有一组特解。
验:令u=3,v=2,k=0;
A=256*uv*(u^4-v^4)^(15k+10)*(u^4+v^4)^(15k+12)=256*6*65^10*97^12
B=32*(u^4-v^4)^(10k+7)*(u^4+v^4)^(10k+8)=32*65^7*97^8
C=8*(u^4-v^4)^(6k+4)*(u^4+v^4)^(6k+5)=8*65^4*97^5
A^4=2^32*6^4*65^40*97^48=2^30*65^40*97^48*(2*36)^2
B^6=2^30*65^42*97^48=2^30*65^40*97^48*65^2
C^10=2^30*65^40*97^50=2^30*65^40*97^48*97^2
72,65,97是一组勾股数,
A^4+B^6=C^10又有一组特解。
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