Square sum

[Guest]

How do I prove that if [tex]a^2 + b^2 \ne c^2 +d^2[/tex] and if [tex]a \ne b[/tex], [tex]c \ne d, a, b, c, d \ge 0[/tex] and [tex]L \ge a,b,c,d[/tex] then [tex](a+L)^2 + (b +L)^2 \ne (c + L)^2 + (d +L)^2[/tex]?

If [tex](a+L)^2 , (b+L)^2, (c+L)^2, (d+L)^2[/tex] are all greater than [tex]a^2,b^2,c^2,d^2[/tex] what is optional L such that it can happen in general the value of L is always greater than the values taken by a,b,c,d in general. If a,b,c,d can take values from 0 to some k-1 for which value of L can we state the above inequallity is true. It is understood counter examples are possible is there any general way to find a suitable L given k and all pairs squares 0 to k-1 allowed

[Guest]

[tex]a=15, b=0, c=10, d=7, L=19[/tex]
satisfies all your constraints, but
[tex](a+L)^2+(b+L)^2=(c+L)^2+(d+L)^2 = 1517[/tex]

Hope this helped,

R. Baber.

[Guest]

The question is contradiction exist finding large L in general such that not equal holds