+0  
 
+4
563
8
avatar+88775 

Can anyone think of 5 positive integers, a,b,c,d and e such that

a^2 + b^2 + c^2 + d^2 = e^2   ?????

[ There is more than one correct answer ]  !!!!!!

 (Also....and I should have been more specific....the integers are all different ) !!!!

CPhill  Oct 8, 2014

Best Answer 

 #4
avatar+5454 
+28

Hm...

a = 4, b = 8, c = 10, b = 12, and e = 18 :)

kitty<3  Oct 10, 2014
 #1
avatar
0

is the answer 1 

Guest Oct 8, 2014
 #2
avatar+26965 
+5

a = b = c = d = n;  e = 2n;  where n is any positive integer.

Alan  Oct 8, 2014
 #3
avatar+93289 
0

Chris said all the integers were different!  

Melody  Oct 9, 2014
 #4
avatar+5454 
+28
Best Answer

Hm...

a = 4, b = 8, c = 10, b = 12, and e = 18 :)

kitty<3  Oct 10, 2014
 #5
avatar+88775 
0

Ding!!  Ding!! Ding!!!......we have a winner!!

kitty<3  !!!!

I came up with........

1 , 2 , 8, 10, 13

Note, that any positive integer multiples of these answers "work," too. (Just like in the Pythagorean "Multiples")

[Mmmm...I wonder if there is a general "formula" for generating the "correct" values??? ]

 

CPhill  Oct 10, 2014
 #6
avatar+26965 
+5

Here's a way of generating more sets of values:

 

Choose a Pythagorean triple, say 3, 4, 5.  Let a = 3m, b = 4m, c = 3n, d = 4n, and e = 5√(m2 + n2

 

a2 + b2 + c2 + d2 = (32 + 42)m2 + (32 + 42)n2 = (32 + 42)(m2 + n2) = 52(m2 + n2) = e2

 

Now choose m and n to be part of another Pythagorean triple, say m = 5 and n = 12, and we have e as an integer (=5*13).

 

So a = 3*5 = 15;  b = 4*5 = 20;  c = 3*12 = 36;  d = 4*12 = 48; e = 5*13 = 65;

$${{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{36}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{48}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

$${{\mathtt{65}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

 

(I don't remember seeing Chris's last line when I gave my previous answer.  Hmm!)

 

.

Alan  Oct 10, 2014
 #7
avatar+88775 
0

Very nice, Alan.....I added that last line after you posted your first answer.....I meant for the all the digits to be different.......

 

CPhill  Oct 10, 2014
 #8
avatar+93289 
0

Well Alan, I shall, give you 3 points for your first answer then - otherwise you would be getting cheated!  

Melody  Oct 11, 2014

11 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.