I have asked this question before, but didn't really understand the answer given. I found this proof elsewhere.


$a^2 \le x < (a + 1)^2$ is true for some (unique) positive integer $a$.

We have to show there is some $y > 0$ such that $ y + x = m^2$ is a perfect square for some $m\in\Bbb N$. Since there are no $y > 0$ such that $y + x=$ perfect square for $x = 1, 2, 3, 4$, we claim that the proposal is true for $x \ge 5$.

If $x = 5, 6, 7, 8$, then $a$ is not an integer, so the inequality above won't hold since we know $a$ is a positive integer. So, we'll check these cases separately.

Suppose $y > 0$.

Then $5 + 4$ is a perfect square. So are $6 + 3$, $7 + 2$ and $8 + 1$.

So it's sufficient to prove our claim for $x > 9 $.

If $x > 9$, then $a \ge 3$.

So $a^2 \le x < (a + 1)^2$ holds.

$a^2 -x \le 0 < (a + 1)^2 -x$ follows from $a^2 \le x < (a + 1)^2$.

Since $(a + 1)^2 -x > 0$, we let $y = (a + 1)^2 -x$.

To justify our choice of $y$ and to prove our claim, we also have to show that $x + y$ is a perfect square and that $x > y$:

$x + y = x + (a + 1)^2 -x = (a + 1)^2$.

$ y = (a + 1)^2 - x$

$ < (a + 1)^2 - a^2$

$= 2a + 1$

$< a^2$

$\le x$.

How can $a = 3$ if $x > 9$? It's obvious that $y$ is derived from $a^2 \le x < (a + 1)^2$, so does that mean that if we prove any statement for $x$ using this particular $y$, then the statement will hold for all $x > 9$? I am just trying to see the relevance of the bolded line in the proof to everything that follows it. At $x = 10$, $a$ is not an integer, so how can we say that our statement is true for all $x > 9$? Very confused here.

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    $\begingroup$ This is ridiculously false. If this were true, then all numbers greater than $x$ would be perfect squares. $\endgroup$ – Adam Hughes Jul 20 '14 at 21:51
  • $\begingroup$ Where can I look up facts about this proposal(theorem)? $\endgroup$ – Erbolat Jul 20 '14 at 21:53
  • $\begingroup$ There are none other than it is completely false. Consdider $x=5$ and $y=1$, $5+1=6$ is not a perfect square. Your first step: $a^2\le x\ < (a+1)^2$ is not true for all positive integers, $a$, let $x=5$ and $a=3$, then $a^2=9>5=x$. $\endgroup$ – Adam Hughes Jul 20 '14 at 21:54
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    $\begingroup$ @AdamHughes If you're here to pick on how he's formulated himself, why don't you propose comcrete improvements? Because it wouldn't take a lot of thinking to figure out what he's really after. Here, I'll go first: When you say "is true for all positive integers $a$", you should really say "is true for some positive integer $a$". $\endgroup$ – Arthur Jul 20 '14 at 21:55
  • $\begingroup$ @Arthur I'm happy to do so, but if the op just has things changed, he may not get what is wrong with what he's doing and so it wouldn't be much of a help to him: half the battle is being able to communicate the right result. In particular there are enough errors and strange assumptions, that I cannot divine what he's actually going for. He says $x\ge 5$ is necessary, which it isn't, so it could be there's a vital, missing hypothesis. $\endgroup$ – Adam Hughes Jul 20 '14 at 21:56

Given the current formulation we proceed as follows:

Because $x\ge 5$ we know that $a\ge 2$ so that $(a+1)^2-a^2 = 2a+1\ge 5$, which gives us $(a+1)^2>9$

Then we can let $y=(a+1)^2-x>0$ by assumption. It remains to check $x>y$, but then

$2a+1\ge y$ and $x\ge a^2$, when $a\ge 3$ this means $a^2\ge 2a+1$, so checking the other cases, $x=5,6,7,8$ by hand and noting $x\ge a^2> 2a+1\ge y$ when $a>2$, the result follows.

In particular, we cannot claim $a=3$, the key is that $a$ is at least 3 when $x\ge 9$, since that makes it so that $a^2>2a+1$ is a strict inequality.


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