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I have a number theory question:

Consider any three consecutive natural numbers the smallest of which is greater than $3$. Then prove that square of largest cannot be the sum of squares of other two.

I took three natural numbers and tried to proceed with given conditions, but my approach is not working.

Please help.

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  • $\begingroup$ Do you think this is applicable on $n<3$ $\endgroup$ – user401699 Jan 16 '17 at 1:20
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Let $n,n+1,n+2$ be three consecutive natural numbers and $n>3$ (as required).

With given condition, let us assume that there exist three such consecutive natural numbers such that:

$(n+2)^2=n^2+(n+1)^2$

$\implies n^2+4n+4-(2n^2+2n+1)=0$

$\implies -n^2+2n+3=0$

$\implies -(n+1)(n-3)=0$.

Notice that the LHS cannot be zero as $n>3$.

A contradiction.

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    $\begingroup$ This is a good example of a proof by contradiction that really isn’t one! Instead of making an assumption that there is a solution and showing it to be impossible, you can directly show that that quadratic equation has no solutions $> 3$. $\endgroup$ – Lynn Dec 26 '16 at 16:53
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    $\begingroup$ @Lynn, there are various methods which a person can adopt. I respect your opinion but I often use proof by contradiction because I like them. I agree that there are various method and I will also upvote an answer having one, but for me this is okay. I would tend to give an answer which I like even if it cost me 2-3 lines more. Thanks for your words. $\endgroup$ – Vidyanshu Mishra Dec 26 '16 at 16:56
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    $\begingroup$ I'd love to try and convince you that "forced" proofs by contradiction like the above are generally a little less helpful than their corresponding direct proofs, but Qiaochu has already written what I wanted to write :) math.stackexchange.com/a/319926/259262 $\endgroup$ – Patrick Stevens Dec 30 '16 at 11:57
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    $\begingroup$ @PatrickStevens, I m glad to have your words and the link you provided. Thanks. You know what, you convinced me. I know that proof by contradiction are a little vague to establish, but as I explained in my previous comment, it is the way I do things. But now, from your and lynn's comment, I have changed my mind (not totally) and I shall keep it in my mind while writing proofs in future. $\endgroup$ – Vidyanshu Mishra Dec 30 '16 at 12:07
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Suppose we have three numbers, $n, n+1, n+2$ where $n > 3$. Suppose that $(n+2)^2 = n^2 + (n+1)^2$ for contradiction. Then $$n^2 + 4n + 4 = 2n^2 + 2n + 1 \implies -n^2 + 2n +3 \implies n = \frac{-2 \pm \sqrt{4 + 4(3)}}{-2}$$So the only cases where the largest number squared is equal to the sum of squares of the two smaller numbers is when $n = -1, 3$, both of which preclude the assumption that $n>3$.

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Consider $(n+2) \times (n+2)$ balls. Count balls at the boundary, in two ways:

enter image description here

The 4 sides contribute $4 \times (n+1)$ balls as shown. On the other hand, the boundary is the total $ (n+2)^2$ minus the balls in the inner square $n^2$. Therefore, $(n+2)^2 - n^2 = 4 (n+1) \neq (n+1)^2$ unless $n = 3$.

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The single counterexample $16+25\ne36$ suffices.

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  • $\begingroup$ Though the quantification in the OP is unclear. $\endgroup$ – Yves Daoust Dec 30 '16 at 11:43
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Suppose that $n^2+(n+1)^2 = (n+2)^2 $, which is true for $n = 3$.

Then, if $k > 0$ and $n \ge 1$,

$\begin{array}\\ (n+k)^2+(n+1+k)^2-(n+2+k)^2 &=n^2+2nk+k^2+(n+1)^2+2k(n+1)+k^2\\ &\quad -((n+2)^2+2k(n+2)+k^2)\\ &=n^2+(n+1)^2-(n+2)^2+2nk+2k(n+1)\\ &\quad-2k(n+2)+k^2+k^2-k^2\\ &=2k(n+(n+1)-(n+2))+k^2\\ &=2k(n-1)+k^2\\ &> 0\\ \end{array} $

so that $(n+k)^2+(n+1+k)^2 >(n+2+k)^2 $.

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There's not so much space between consecutive squares! $$(n+2)^2=(n+1+1)^2=(n+1)^2+2(n+1)+1<(n+1)^2+n^2$$ as soon as $2n+3<n^2$, i.e. $(n-1)^2>4$.

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$$(n+2)^2-n^2= [(n+2)-n]\cdot[(n+2)+n]= 2(2n+2)=4(n+1) $$

So if you want $4(n+1)=(n+1)^2$, you have either $n+1=0$ or $n+1=4$.

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