Proving Divisibility of x, y and z by 5 in Modulo 5 Problem

[doggie_Walkes]

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 I am not sure if I am 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 don't know wher to go. can someone help me, or show me how to prove this.

would be greatly appreicated.

thanks

 

[tiny-tim]

Hi doggie_Walkes!

… 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 … ?

 

[doggie_Walkes]

thanks tiny tim for replying so quickly.

well I am 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

 

[tiny-tim]

Hi doggie_Walkes!

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

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

 

[doggie_Walkes]

Ah tiny tim, I get it! thanks that's 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

 

[tiny-tim]

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! ​

 

[doggie_Walkes]

I still don't get it :(

 

[tiny-tim]

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 …