# Prove perfect squares

• May 19th 2010, 10:10 AM
Mukilab
Prove perfect squares
$\displaystyle \sqrt{p} +\sqrt{q} +\sqrt{z} =s$

Prove that p, q and r must be perfect squares.

Ok so the rule for a square is: $\displaystyle n^2=(n-1)^2+\sqrt{(2n-1)}$

So $\displaystyle p=p-1+\sqrt{2p-1}$ then $\displaystyle p-p=-1+\sqrt{(2p-1)}$ therefore if you square it $\displaystyle 0=1+2p-1$ which goes to $\displaystyle 0=2p$

Where did I go wrong?
• May 19th 2010, 11:04 AM
Pim
I think you went wrong in your first step.

If you substitute $\displaystyle n = \sqrt{p}$ into $\displaystyle n^2 = (n-1)^2 + \sqrt{2n-1}$ you get:

$\displaystyle p = (\sqrt{p}-1)^2 + \sqrt{2 \sqrt{p} - 1}$ and not what you wrote

Another thing is that

$\displaystyle (-1 + \sqrt{2p-1})^2 \neq 1 + 2p-1$
it doest equal $\displaystyle (-1 + \sqrt{2p-1})(-1 + \sqrt{2p-1})$, where you multiply out the brackets.
• May 19th 2010, 12:11 PM
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Quote:

Originally Posted by Mukilab
Ok so the rule for a square is: $\displaystyle n^2=(n-1)^2+\sqrt{(2n-1)}$

Where did you get this? If you plug in $\displaystyle n=2$, the right hand side yields $\displaystyle 1+\sqrt{3}$ instead of $\displaystyle 4$.

From context it's clear that you mean $\displaystyle p, q, r, s \in \mathbb{Z}$.

Then it's a matter of proving that if p, q and r are not all perfect squares, then s cannot be rational.
• May 19th 2010, 01:11 PM
Mukilab
That's the formula for proving the square of a number using the previous number squared.
• May 19th 2010, 01:12 PM
Mukilab
I don't know if it's a matter of whether p, q and r are perfect squares or not.

Never mind, this question does not apply to me anymore and I have far more tricky questions to get around :P
• May 19th 2010, 01:16 PM
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Quote:

Originally Posted by Mukilab
That's the formula for proving the square of a number using the previous number squared.

Well just for reference,

$\displaystyle (n-1)^2=n^2-2n+1$

$\displaystyle n^2=(n-1)^2+2n-1$