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Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.

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    $\begingroup$ This task has already been!!!! math.stackexchange.com/questions/2126403/mohantys-conjecture $\endgroup$ – individ Mar 3 '17 at 17:48
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    $\begingroup$ Umm, I think my question is a bit different $\endgroup$ – user421996 Mar 3 '17 at 17:51
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    $\begingroup$ @individ I don't see any resemblance between this question and that one. $\endgroup$ – lulu Mar 3 '17 at 17:52
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    $\begingroup$ @J.Doe It is different, clearly. not to worry. $\endgroup$ – The Count Mar 3 '17 at 17:53
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    $\begingroup$ What else? Difficult $T=1$? $\endgroup$ – individ Mar 3 '17 at 18:01
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n = 0 is a trivial solution as 0 = $0^2 + 0^2, 1 = 1^2 + 0^2, 2 = 1^2 + 1^2$. Consider the pell equation $x^2 - 2y^2 = 1$, which has infinitely many solutions. Take $n = x^2 -1$. This implies $ n = y^2 + y^2$, $n+1 = x^2 + 0^2$ and $n+2 = x^2 + 1^2$.

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One such infinite family is $$(2a^2 + 1)^2 - 1, (2a^2 + 1)^2 + 0^2, (2a^2 + 1)^2 + 1^2$$ for all integer $a$. This works because $$(2a^2 + 1)^2 - 1 = 4a^4 + 4a^2 = (2a^2)^2 + (2a)^2.$$

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    $\begingroup$ This should be the accepted answer! $\endgroup$ – N.S.JOHN Mar 4 '17 at 8:07
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If we can find $y$ such that $y^2-1$ is a sum of two squares, say $y^2- 1 = a^2+b^2$, then take $n = y^2-1$ and we'll have $n = y^2-1 = a^2+b^2$, $n+1 = y^2-1 +1 = y^2 +0$, $n+2 = y^2 + 1$. So, we want to show the existence of infinitely many $y$ such that $y^2-1$ is a sum of two squares.

Look at $3^{2^k}$ for integers $k \geq 1$. For $k = 1$, $3^2 - 1= 8 = 2^2 + 2^2$.

Now, $3^{2^2} -1 = 3^4 -1 = (3^2-1)(3^2+1) = (2^2 +2^2)(3^2+1^2) $

Note the identity $(p^2+q^2)(c^2+d^2) = (pc+qd)^2 + (pd-qc)^2$. So we know that the product of two sums of two squares is again a sum of two squares. And hence, without explicitly calculating, we can say $3^{2^2}-1$ is a sum of two squares.

Now suppose for the sake of induction that $3^{2^k} - 1$ is a sum of two squares. Then $3^{2^{k+1}} -1 = (3^{2^k}-1)((3^{2^{k-1}})^2 +1^2)$. Thus from the induction hypothesis and our neat identity, $3^{2^{k+1}} -1$ is a sum of two squares.

So, for all integers $k \geq 1$, taking $y = 3^{2^k}$ and $n = y^2-1$gives us infinitely many integers satisfying the problem's requirement.

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I think it's better to solve the problem for the General case. For the case of even numbers the solution there. Mohanty's conjecture

For odd numbers.

$$\left\{\begin{aligned}&N=a^2+b^2\\&N+T=c^2+q^2\\&N+2T=x^2+y^2\end{aligned}\right.$$

Decompose the number into factors. $T=ps$

The solution can be written as.

$$a=pk^2+(2p+s)k+\frac{p+3s}{2}$$

$$b=pk^2+(2p+s)k+\frac{3p+s}{2}$$

$$c=pk^2+(3p+s)k+\frac{3(p+s)}{2}$$

$$q=pk^2+(p+s)k-\frac{p-s}{2}$$

$$x=pk^2+(2p+s)k+\frac{p+s}{2}$$

$$y=pk^2+(2p+s)k+\frac{3(p+s)}{2}$$

It is seen that for all odd numbers are infinitely many solutions, not just $1$.

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    $\begingroup$ Now that's a great answer! Unlike all others, it is not hinged on using 0 as one of the squares. However, I'd rather write expressly the case of $p=s=1$ (anyway, this is the only one we're talking about), so that people won't have to do the calculations themselves. In short, our numbers are: $$n=2(k^2+3k+2)^2 \\ n+1=(k^2+4k+3)^2+(k^2+2k)^2 \\ n+2=(k^2+3k+1)^2+(k^2+3k+3)^2$$ $\endgroup$ – Ivan Neretin Mar 10 '17 at 6:42

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