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Modulo 5 problem

  • #1
Well the problem is

Prove that if x, y, z are intergers such that 5*x^2 + y^2 = 7*z^2, then x, y and z are all divisble by 5.


So what I have done so far,

I have worked out 1, 2 ,3 , 4, and their squared to find that. the squared intergers of any interger will end in 0,1, 4 in modulo 5. (ps im not sure if im phrasing this write as well)

then the LHS would be 0, 1,4. whilst the right hand side will be 0, 2, 3.

now i dont know wher to go. can someone help me, or show me how to prove this.

would be greatly appreicated.

thanks
 

Answers and Replies

  • #2
tiny-tim
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Hi doggie_Walkes! :wink:
… then the LHS would be 0, 1,4. whilst the right hand side will be 0, 2, 3.
Yes! So both sides must be 0, and so … ? :smile:
 
  • #3


thanks tiny tim for replying so quickly.

well im checked the answer then its says this,

"the only possiblity is that y=z=0(mod5)

but then 5*x^2=7*z^2 - y^2 is divisble by 25, and so x is divisble by 5."



so i get why the obly possiblity is 0mod5 but why is it divisble by 25.

regards
 
  • #4
tiny-tim
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Hi doggie_Walkes! :smile:

(try using the X2 tag just above the Reply box :wink:)

ah, because if 5|y and 5|z, then 25|y2 and 25|z2 (and so 25|5x2) :wink:
 
  • #5


Ah tiny tim, I get it! thanks thats bothering for some time, can i ask one more thing of you please. How would one go about doing this?

5x2+y2 = 7z2

Deduce that the equation has no solution in intergers except fo x=y=z= 0
 
  • #6
tiny-tim
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Deduce that the equation has no solution in intergers except fo x=y=z= 0
but that follows directly from the first result …

think about it! :smile:
 
  • #7


I still dont get it :(
 
  • #8
tiny-tim
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5 divides x y and z, so put x = 5a, y = 5b, z = 5c, then 5a2 + b2 = 7c2.

Now 5 divides a b and c, so put a = 5p, b = 5q, c = 5r, and so on … :smile:
 

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