Square sum

vs140580

New member
Joined
Nov 15, 2015
Messages
3
How do I prove the following?

If

. . .\(\displaystyle \mbox{i) }\,a^2 \,+\, b^2\, \ne\, c^2\, +\,d^2\,\)

. . .\(\displaystyle \mbox{ii) }\,a\, \ne\, b\,\mbox{ and }\, c\, \ne\, d\)

. . .\(\displaystyle \mbox{iii) }\,a,\, b,\, c,\, d\, \ge\, 0\,\)

. . .\(\displaystyle \mbox{iv) }\,L\, \ge\, a,\, b,\, c,\, d,\)

then \(\displaystyle \,(a\,+\,L)^2 \,+\, (b\, +\,L)^2\, \ne\, (c\, +\, L)^2 \,+\, (d\, +\,L)^2\)



If \(\displaystyle (a+L)^2 , (b+L)^2, (c+L)^2, (d+L)^2\) are all greater than \(\displaystyle a^2,b^2,c^2,d^2\) 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
 
Last edited by a moderator:
How do I prove the following?

If

. . .\(\displaystyle \mbox{i) }\,a^2 \,+\, b^2\, \ne\, c^2\, +\,d^2\,\)

. . .\(\displaystyle \mbox{ii) }\,a\, \ne\, b\,\mbox{ and }\, c\, \ne\, d\)

. . .\(\displaystyle \mbox{iii) }\,a,\, b,\, c,\, d\, \ge\, 0\,\)

. . .\(\displaystyle \mbox{iv) }\,L\, \ge\, a,\, b,\, c,\, d,\)

then \(\displaystyle \,(a\,+\,L)^2 \,+\, (b\, +\,L)^2\, \ne\, (c\, +\, L)^2 \,+\, (d\, +\,L)^2\)



If \(\displaystyle (a+L)^2 , (b+L)^2, (c+L)^2, (d+L)^2\) are all greater than \(\displaystyle a^2,b^2,c^2,d^2\) 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
Do you have an objection to the replies posted here? Please be specific. Thank you! ;)
 
Kind help square sum inequallity

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

If $(a+L)^2 , (b+L)^2, (c+L)^2, (d+L)^2$ are all greater than $a^2,b^2,c^2,d^2$ 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. Thank you sir kind help with an optimal choice of L kind help
Thank you kind help
 
Top