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How many positive integers $x$ are there so that $\sqrt{144+x^2}$ is an integer?

Attempt a solution: I used trial and error and only got $x=5$

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    $\begingroup$ Rephrased, find all pythagorean triples involving $12$ as one of the two smaller entries. There are several including $(5,12,13),(9,12,15),(12,16,20),(12,35,37)$. $\endgroup$ – JMoravitz Feb 20 '18 at 16:17
  • $\begingroup$ Hint, they must lie in $(-\infty, -12] \cup [12, +\infty )$ :-) $\endgroup$ – Kevin Feb 20 '18 at 16:21
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    $\begingroup$ Also, you could re-write to assume $\exists\, z = \sqrt{(12-x)(12+x)}$ $\endgroup$ – Kevin Feb 20 '18 at 16:21
  • $\begingroup$ I completed my answer—no more constant edits now :) $\endgroup$ – Feeds Feb 20 '18 at 17:08
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    $\begingroup$ @JMoravitz those are the only $4$ solutions, actually :) $\endgroup$ – Feeds Feb 20 '18 at 17:15
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You have $$y^2=12^2+x^2\iff (y-x)(y+x)=2^4\cdot3^2$$ The integer $144$ has $(4+1)(2+1)=15$ factors so one can looking at the systems $$\begin{cases}x+y=a\\x-y=b\end{cases}$$ where $$(a,b)\in\{(144,1),(72,2),(48,3),(36,4),(24,6),(18,8),(16,9),(12,12)\}$$ from which we get four solutions $$(a,b)\in\{(72,2),(36,4),(24,6),(18,8)\}$$ giving $$\color{red}{(y,x)\in\{(37,35),(20,16),(15,9),13,5)\}}$$

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    $\begingroup$ $$(y, x)\neq (37, 15)$$ $\endgroup$ – Feeds Feb 20 '18 at 16:50
  • $\begingroup$ I have edited, Thanks you very much ( $37^2-15^2=1144$ and not $144$ $\endgroup$ – Piquito Feb 20 '18 at 16:54
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    $\begingroup$ Hahah no problem at all :) ..... Oh, and $x = 35$ is a solution. $\endgroup$ – Feeds Feb 20 '18 at 16:56
  • $\begingroup$ Very kind. Thanks again. $\endgroup$ – Piquito Feb 20 '18 at 16:57
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    $\begingroup$ It was a lapsus with $15$ instead of $35$. Hahah, as you say. $\endgroup$ – Piquito Feb 20 '18 at 17:01
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substituting $$y=\sqrt{144+x^2}$$ we get by squaring and factorizing $$(y-x)(y+x)=144$$ can you finish?

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Well then since $144 = 12^2$ then you want to find values of $x$ to satisfy the equation, $$12^2 + x^2 = y^2.$$ Now since $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$ then when we multiply them together, we get that $$(a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2.\tag1$$ What you have done to solve for $x$ is by letting $12 = 2ab$ and thus $a = 2$ and $b = 3$. We now want $12 = a^2 - b^2 = (a + b)(a - b)$. Now what divides into $12$? Well the answers are $1, 2, 3, 4, 6, 12$.

Let $a + b = 12$, then $a - b = 1$, therefore $a = b + 1$ and $2b + 1 = 12$. That is a contradiction because $a$ and $b$ are integers if $2ab$ could equal $12$ in the first place, and $12$ is even, so we have to rule that out.

Let $a + b = 6$ then $a - b = 2$, Therefore $a = b + 2$ and $2b + 2 = 2(b + 1) = 6$. Therefore, $b = 2$ and thus $a = 4$.

We have now found another solution for $x$, such that by letting $x = 2ab$ instead (since we let $12 = a^2 - b^2$), we have that $x = 2\times 2\times 4 = 16$.

Now let $a + b = 4$ then $a - b = 3$, Therefore $a = b + 3$ and $2b + 3 = 4$. That is a contradiction, so we rule that out. Now we can’t let $a + b < 4$ because otherwise $a + b < a - b$ which is also a contradiction, so we stop there.

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Now in Eq. $(1)$, if we let $b = 1$ then $$(a^2 - 1)^2 + (2a)^2 = (a^2 + 1)^2.$$ By letting $12 = 2a$ since $12 + 1 = 13 \neq$ a squared number, then we get that $a = 6$. By letting $x = a^2 - 1$ then we obtain another solution for $x$, namely $x = 6^2 - 1 = 36 - 1 = 35.$

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Now it is also provable that for any integer $n$, $$(2n^2 + 2n)^2 + (2n + 1)^2 = (2n^2 + 2n + 1)^2.$$ Since $12$ is even, then $12\neq 2n + 1$ and therefore $12 = 2n^2 + 2n$. We now have the following quadratic trinomial to solve: $$2n^2 + 2n - 12 = 0,$$ and using the quadratic formula, we get that $$n = \frac{-2\pm \sqrt{4 + 96}}{2} = \frac{-2}{2} \pm \frac{10}{2} = -1\pm 5.$$ Therefore, $n = 4$ or $n = -6$. Now, we let $x = 2n + 1$ for those values of $n$ and we get two more solutions for $x$, namely $x = 9$ and $x = -11$.

However, $x > 0$ because we established earlier that if $x = 2ab$ then $a + b\not< a - b$, and if $x < 0$ then either $a$ or $b$ is negative which arouses a contradiction in the strict inequality, so we get only one new solution for $x$ such that $x = 9$. (This also proves why $x$ must be a positive integer.)

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Overall, our solutions for $x$ so far are $x = 5, 9, 16, 35$.


Checking the equation $u^2 + v^2 = w^2$ for $w \leqslant 10,000$, there does not seem to be another solution where either $u$ or $v$ equals $12$ with the other remaining variable on the LHS not equal to any of our listed values of $x$. From this, I conjecture that there only exists $4$ such solutions of $x$.


Main Answer:

According to @Piquito’s answer below $\downarrow\downarrow\downarrow$ there exist only $4$ solutions, and thus the conjecture is valid. I apologise for constantly editing my answer, but I did not have a big calculator with me. I was able to do the calculation concerning $u^2 + v^2 = w^2$ from a link in one of the comments. This meant that I had to find the values of $x$ with pure maths, and I was not going to sit there and exhaustively find solutions for $(x, y)$ in the equation, $144 = (y + x)(y - x)$.

Consequently, I decided to be creative.

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$$(x+2a)^2=x^2+144$$ where $a>0$

$$144=4a(x+a)>4a^2\implies a<6$$

and $a$ divides $36$

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  • $\begingroup$ Why is it (x+2a)? $\endgroup$ – Janjan Feb 20 '18 at 16:35
  • $\begingroup$ @Janjan, Start with $$(x+b)^2=x^2+144$$ $\endgroup$ – lab bhattacharjee Feb 20 '18 at 16:40

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