Proof

[keylargo]

If someone could point me in the correct direction to start this proof I would be most appreciative:

Prove there does not exist 2 integers a and d with a^2 - 8d^2 = 35 using this information:
1. use contradiction
2. consider the integers modulo 8
3. use modular arithmetic
4. use the result of the division algorithm that every integer is congruent modulo m to exactly 1,2,3,..,m-1

 

[galactus]

Consider the inverse of 8 modulo 35.

 

[keylargo]

the inverse of 8 modulo 35 is -13
does this have to do with the 4th bit of information given?

 

[lookagain]

If someone could point me in the correct direction to start this proof I would be most appreciative:

Prove there does not exist 2 integers a and d with a^2 - 8d^2 = 35 using this information:
1. use contradiction

What if this proof used the first condition alone?

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Let \(\displaystyle M\) be an integer.

Assume there do exist integers \(\displaystyle a \ and \ d\) such that \(\displaystyle a^2 - 8d^2 = 35.\)

Then \(\displaystyle a^2 = 8d^2 + 35.\)

And \(\displaystyle a^2 = 8d^2 + 32 + 3.\)

\(\displaystyle a^2 = 4(2d^2 + 8) + 3\)

\(\displaystyle Then \ a^2 = 4M + 3.\)

But the square of an integer can only be of the form \(\displaystyle 4M \ or \ of \ the \ form \ (4M + 1).\)

Because \(\displaystyle a^2\) is not of this form, then our original assumption is false,
and there are no integers \(\displaystyle a \ and \ d\) with that constraint.

 

[sgreany]

i have to prove this to be true or show a counter example if its false:

if x is rational then x^2/(1+x^2) is rational

 

[Mrspi]

i have to prove this to be true or show a counter example if its false:

if x is rational then x^2/(1+x^2) is rational

Next time, please start your own thread by clicking on "New Topic," rather than hijacking someone else's thread.

That said,

you are GIVEN that x is rational. By definition, then, x = a/b where a and b are integers and b is NOT 0.

Substitute a/b for x in

x[sup:2rgji4cu]2[/sup:2rgji4cu] / (1 + x[sup:2rgji4cu]2[/sup:2rgji4cu])

(a/b)[sup:2rgji4cu]2[/sup:2rgji4cu] / [1 + (a/b)[sup:2rgji4cu]2[/sup:2rgji4cu]]

Now, play around with that a bit, and see if you can get it in the form of a simple fraction which has an integer for its numerator and for its denominator (should not be too difficult...you already KNOW something about a and b)