Solve and analyze the problem as follows: Let these four numbers be x,y,z,m and satisfy:
x+y=a^2 .... ①
x+z=b^2 .... ②
y+z=c^2 .... ③
x+m=d^2 .... ④
y+m=e^2 .... ⑤
z+m=f^2 .... ⑥
(!!!! !!!!) The middle is the comment part, for easy understanding of the input, may not look
In fact, which only x,y,z,w four unknowns (a,b,c,d,e,f is known), but there are six equations, which a,b,c,d,e,f must have a special relationship, (comment!!!!). For example, x+y=2,x-y=0, 2x+y=m For this system of equations to have a solution, it must be m=3 Another example, x+y=2p,x-y=2q,2x+y=m For this equation to have a solution, from the first two equations, we have x=p+q,y=p-q,Substituting the third one, we can find out the relationship between m and p,q. m=3p !!!! (End of note)
And in this question it is clear that , according to . ①,②,③ we can solve for x,y,z (denoted by a^2,b^2,c^2) and ④. ⑤. ⑤. ⑥ three equations, but only one m left, so we can get the relationship between a,b,c,d,e,f according to these three equations, and further find a,b,c,d,e,f and then x,y,z,m
From ① + ⑥, we get m=a^2+f^2-x-y-z
From ② ⑤, we get m=b^2+e^2-x-y-z
From ③④, m=c^2+d^2-x-y-z
According to the above three equations, we get a^2+f^2=b^2+e^2=c^2+d^2, which means that a,b,c,d,e,f satisfy the relationship
The question then transforms into: Is there a situation where a,b,c,d,e,f satisfy a^2+f^2=b^2+e^2=c^2+d^2+c^2+e^2? =c^2+d^2 holds.
Here is an idea to find such numbers,
First find three sets of Pythagorean numbers 3^2+4^2 =5^2 , 5^2+12^2=13^2, 8^2+15^2=17^2 and then expand it by a certain number of times, so that the three sets of hypotenuse squares are equal
3^2+4^2 =5^2 Multiply the two sides by (13*17)^2 and you get:(3^2+4^2=5^2)^2 We get:(3*13*17)^2+(4*13*17)^2=(5*13*17)^2
5^2+12^2=13^2, multiply both sides by (5*17)^2 and we get:(5*5*17)^2+(12*5*17)^2=(5*13*17)^2
Similarly 8^2+15^2=17^2. Multiply both sides by (5*13)^2 to get, (8*5*13)^2+(15*5*13)^2=(5*13*17)^2
So we get the answer that meets the condition:
(3*13*17)^2+(4*13*17)^2=(5*5*17)^2+(12*5*17)^2=(8*5*13) ^2+(15*5*13)^2
That is a=3*13*17=663, f=4*13*17=884, b=5*5*17=425 ,e=12*5*17=1020,c=8*5*13=520,d=15*5*13=975
You can get a^2=439569 ,b^ 2=180625 ......
Substitute ①, ②, ③, ④, ⑤, ⑥ to solve the answer
x=174897 y=264672 z=5728 m=775728
(You can also substitute ①, ②, ③ to solve for x,y,z and then solve for m)
It took me a long time to type it in, and it's a good question!