Question #276264

List the quadruples in the relation { a,b,c,d} where a, b, c, d are integers with

0 < a < b < c < d < 8


1
Expert's answer
2021-12-07T06:49:15-0500

A Pythagorean quadruple is a tuple of integers a,b,c,a, b, c, and d,d, such that a2+b2+c2=d2.a^2 + b^2 + c^2 = d^2.


12+22+32=14,No1^2+2^2+3^2=14, No

12+22+42=21,No1^2+2^2+4^2=21, No

12+22+52=30,No1^2+2^2+5^2=30, No

12+22+62=41,No1^2+2^2+6^2=41, No

12+32+42=26,No1^2+3^2+4^2=26, No

12+32+52=35,No1^2+3^2+5^2=35, No

12+32+62=46,No1^2+3^2+6^2=46, No

12+42+52=42,No1^2+4^2+5^2=42, No

12+42+62=53,No1^2+4^2+6^2=53, No

12+52+62=62,No1^2+5^2+6^2=62, No

12+62+72=86,No1^2+6^2+7^2=86, No

22+32+42=29,No2^2+3^2+4^2=29, No

22+32+52=38,No2^2+3^2+5^2=38, No

22+32+62=49=72,Yes2^2+3^2+6^2=49=7^2, Yes

22+42+52=45,No2^2+4^2+5^2=45, No

22+42+62=56,No2^2+4^2+6^2=56, No

22+52+62=65,No2^2+5^2+6^2=65, No

32+42+52=50,No3^2+4^2+5^2=50, No

32+42+62=61,No3^2+4^2+6^2=61, No

32+52+62=70,No3^2+5^2+6^2=70, No

42+52+62=77,No4^2+5^2+6^2=77, No



{2,3,6,7}\{2,3,6,7\}


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