Consecutive squares

Note that these squares form a primitive Pythagorean Triplet, so there exist integers $p$ and $q$ such that $n = p^2 - q^2$ and $n+1 = 2pq$ (or the other way around). Thus, $p^2 - q^2 = 2pq \pm 1 \implies (p-q)^2 + 2q^2 = \pm 1$ .

Let $u=p-q$ , and $v=q$ , so we have the Pell equations $u^2 - 2v^2 = 1$ and $u^2 - 2v^2 = -1$ .

Part 1: $u^2 - 2v^2 = 1$

The fundamental solution of this can be observed as being $(3, 2)$ . Thus, using the recursive formula $(u_{n+1}, v_{n+1}) = (u_0 u_n + D v_0 v_n, u_0 v_n + v_0 u_n)$ , we get the following solutions: $(3, 2), (17, 12), (99, 70), \ldots$ . These give the respective values for $(p,q)$ : $(5, 2), (29, 12), (148, 70), \ldots$ , which respectively give these values for $n$ : $n=20, 696, 23660, \ldots$ . Thus, all satisfying $n<2017$ in this case are $n=20, 696$ .

Part 2: $u^2 - 2v^2 = -1$

The fundamental solution for this can be observed as being $(1, 1)$ . Thus, using the recursive formula $(u_{n+1}, v_{n+1}) = (u'_0 u_n + D v'_0 v_n, u'_0 v_n + v'_0 u_n)$ , where $(u'_0, v'_0)$ is the fundamental solution of $u^2 - 2v^2 = 1$ , we get the following solutions: $(1, 1), (7, 5), (41, 29), \ldots$ . These give the respective values for $(p,q)$ : $(2, 1), (12, 5), (70, 29), \ldots$ , which respectively give these values for $n$ : $n=3, 119, 4059, \ldots$ . Thus, all satisfying $n<2017$ in this case are $n=3, 119$ .

Thus, we have all satisfying values of $n$ to be $3, 20, 119, 696$ , so their sum is $\boxed{838}$ .

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Note that another solution exists, using Vieta Root Jumping on $(p, q)$ .

We want $n^2 + (n+1)^2 = m^2$ , so that $(2n+1)^2 + 1 = 2m^2$ , and hence $(2n+1)^2 - 2m^2 \; = \; -1$ The positive integer solutions of the equation $x^2 - 2y^2 = -1$ are well-known to be given by the formula $x_k + y_k\sqrt{2} \; = \; (1 + \sqrt{2})^{2k+1} \hspace{2cm} k \ge 0$ (the even powers of $1 + \sqrt{2}$ give the positive integer solutions of $x^2 - 2y^2 = 1$ ). The first $6$ pairs $(x_k,y_k)$ are $(1,1)$ , $(7,5)$ , $(41,29)$ , $(239,169)$ , $(1393,985)$ and $(8119,5741)$ , giving values of $0,3,20,119,696,4059$ for $n$ .

Larger values of $k$ yield larger values of $n$ . Thus the answer is $3 + 20 + 119 + 696 = \boxed{838}$

This is considered as cheating.

Computational approach incoming, now with easy excel that doesn't even need any coding basics.

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Write this in excel.

This is how it works.

Column B would be straightforward to understand, and so is column C;

Now floor it down by column D.

If the result of B is a square, then C is a natural number, then C=D. If the result of B wasn't, then C is not a natural number, then C>D.

Column E tries to calculate the square root of D-C, however, this will return an error: "#NUM!" if C>D.

Therefore, only two values can fill the column E, which are "0"s and "#NUM"s.

Get some filter on top of column E, and you're ready to go.

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Now extend the list as $1\le n \le 2017.$

(Click and drag the bottomright corner of A2 downwards till A2018. Select B2~E2, and click and drag the bottomright corner of E2 downwards till B2018~E2018.)

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And then, click on the filter and deselect "#NUM".

It seems like the answer is $3+20+119+696=\boxed{838}.$

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Furthermore, typing this sequence up in Wolframalpha,

OEIS A001652 is the sequence where ${a_n}^2+(a_{n}+1)^2$ is a square.

The terms are found by a recurrence formula which follows $a_{n+2}=6a_{n+1}-a_{n}+2,~a_0=0,~a_1=3.$

Again, this is cheating, considering this problem is a Number Theory problem.

I do it with C program.

First I keep the perfect squares in an array. Then I input $n$ 's value and check that for which $n$ ans matches with the perfect square which I kept in an array before. After that I added all $n$ for answer.

$(n)^2 +(n+1)^2 = (n+x)^2$ So on expanding we get $n^2 -2n(x-1) - (x^2 -1) = 0$ . So by quadratic equation formula we get $n = \frac{2(x-1) + \sqrt{(2(x-1))^2 +4(x^2 -1)}}{2}$ . For $x$ need to be an integer we need $(2(x-1))^2 +4(x^2 -1)$ to be perfect square. So by expanding we get $(2(x-1))^2 +4(x^2 -1) = 8x(x-1)$ . we can check for perfect square in $2x(x-1)$ . $x$ and $2(x-1)$ needs to be perfect squares or $2x$ and $(x-1)$ needs to be perfect squares. We can check each of the squares doubles. Thus we get 2,8,18,32,50,72,98,128,162,200,242,288,338,392,450,512,578,648,722,800,882,968,1058 etc for first few squares. we get some squares from its +1 or -1 then we get a value. 2-1=1,8+1=9,50-1=49, 288+1=289 are the squares we get from these. substituting these values in the quadratic equation we get $n$ as $3,20,119,696$ . Thus the final answer is $696+119+20+3=838$

My solution wasn't as good as the others. I just used the PPT formulas to set up the two quadratic equations. The discriminant of each of those equations was $2p^2+1$ . So I checked all the values of $p$ from 1 to 40 (by hand too) to see which produced a perfect square. There were only two: $2$ and $12$ . There turned out to be two values of $n$ corresponding to each for a total of four values: $3,20,120,696$ .