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Square

 Lemon Quarter
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Re: Square
That looks more like a problem in lateral thinking than anything mathematical. As in, what is your ellipsis?
In the absence of ellipsis, it just evaluates as 121 + 8n being a square, which is a rather dull problem: for every odd x > 11, let y = (x11) / 2. Then x2  121 = (x11) (x+11) = 2y * (2y+22) = 4y(y+11), which is always divisible by 8. So the values of n are the square roots of every (x2121) for odd values of x starting at 13. That gives us n=6,13,21,30, ...
I haven't bothered with a spoiler 'cos, being one of yours, I'm sure it's not supposed to be that straightforward.
In the absence of ellipsis, it just evaluates as 121 + 8n being a square, which is a rather dull problem: for every odd x > 11, let y = (x11) / 2. Then x2  121 = (x11) (x+11) = 2y * (2y+22) = 4y(y+11), which is always divisible by 8. So the values of n are the square roots of every (x2121) for odd values of x starting at 13. That gives us n=6,13,21,30, ...
I haven't bothered with a spoiler 'cos, being one of yours, I'm sure it's not supposed to be that straightforward.

 Lemon Quarter
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Re: Square
cinelli wrote:A mathematical problem. Find all values of integer n > 1 for which the sum
25 + 38 + 41 + ... + (8n+17)
is a perfect square.
Cinelli
Presumably a small typo there, should read:
A mathematical problem. Find all values of integer n > 1 for which the sum
25 + 33 + 41 + ... + (8n+17)
is a perfect square.

 Lemon Quarter
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Re: Square
GoSeigen wrote:A mathematical problem. Find all values of integer n > 1 for which the sum
25 + 33 + 41 + ... + (8n+17)
is a perfect square.
That at least makes a more obvious problem. Your series reduces to 17*n + 8 * n * (n+1) /2, or more simply n(4n+21). There's one obvious solution to that, where both n and (4n+21) are squares. But I don't instantly see a proof of uniqueness, and both factors being squares is a stronger condition than you've asked.

 Lemon Slice
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Re: Square
GoSeigen wrote:
Presumably a small typo there, should read:
25 + 33 + 41 + ... + (8n+17)
is a perfect square.
You are right  it should be 33. Apologies.
Cinelli

 Lemon Slice
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Re: Square
Spoiler
As noted by Uncle E
25 + 33 + 41 +... + (8n + 17) = n(4n + 21)
A bit of brute force reveals the first 2 solutions occur at n=1 (sum is 25) and at 25 (sum is 3025, square of 55). So is there a pattern here? Trying n as 3025, yields a sum of 36,666,025, but this is not a perfect square. So no pattern based on generating a new solution from the preceding one.
However, looking at the next highest square number for n in the range 4 to 25 does reveal something useful. The next highest square greater than or equal to n(4n + 21) is always the square of s = 2n+5.
(2n + 5)(2n + 5)  n(4n + 21) = 25  n
which is positive for n less than 24, 0 when n is 25 and negative for n greater than 25.
It turns out that the square of s + 1 is always greater than n(4n + 21) since
(s + 1)(s + 1)  n(4n + 21) = 36 + 3n
Since
s^2 < n(4n + 21) < (s + 1)^2
for all n > 25 then for this same range, n(4n + 21) cannot be a perfect square. Therefore, since there are two solutions for n<=25, these must be the only solutions.
As noted by Uncle E
25 + 33 + 41 +... + (8n + 17) = n(4n + 21)
A bit of brute force reveals the first 2 solutions occur at n=1 (sum is 25) and at 25 (sum is 3025, square of 55). So is there a pattern here? Trying n as 3025, yields a sum of 36,666,025, but this is not a perfect square. So no pattern based on generating a new solution from the preceding one.
However, looking at the next highest square number for n in the range 4 to 25 does reveal something useful. The next highest square greater than or equal to n(4n + 21) is always the square of s = 2n+5.
(2n + 5)(2n + 5)  n(4n + 21) = 25  n
which is positive for n less than 24, 0 when n is 25 and negative for n greater than 25.
It turns out that the square of s + 1 is always greater than n(4n + 21) since
(s + 1)(s + 1)  n(4n + 21) = 36 + 3n
Since
s^2 < n(4n + 21) < (s + 1)^2
for all n > 25 then for this same range, n(4n + 21) cannot be a perfect square. Therefore, since there are two solutions for n<=25, these must be the only solutions.

 Lemon Slice
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Re: Square
Reply to spoiler:
Well solved modellingman. It is satisfying when two different methods produce the same answer. I will give my method in a couple of days.
Cinelli
Well solved modellingman. It is satisfying when two different methods produce the same answer. I will give my method in a couple of days.
Cinelli

 Lemon Quarter
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Re: Square
cinelli wrote: I will give my method in a couple of days.
Cinelli
Is your method to enumerate candidate square roots?
The square root of sq = 4n2 + 21n is something above 2n, let's say 2n+x. There are few enough cases to enumerate:
For x <= 2, (2n+x)2 <= 4n2 + 8n + 4 < sq
For x >= 6, (2n+x)2 >= 4n2 + 24n + 36 > sq
That leaves us cases 3, 4 and 5 as candidates.
3: 4n2 + 12n + 9
4: 4n2 + 16n + 16
5: 4n2 + 20n + 25
But we can reduce this to trivial linear equations by subtracting 4n2 from the squares and from sq:
12n + 9 = 21n has solution 1
16n + 16 = 21n has noninteger solution
20n + 25 = 21n has solution 25
So 1 and 25 are indeed the only solutions.

 Lemon Slice
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Re: Square
Here is my solution:
Sum of the AP is n*(4n+21) so we are looking for positive integer k for which
4n^2+21*n = k^2
Multiply through by 16 to avoid fractions: 64*n^2 + 16*21*n  16*k^2 = 0
Complete square: (8n+21)^2 – 441 – 16*k^2=0
Factorise difference of two squares: ((8n+21)+4k) * ((8n+21)4k) = 21^2
Possible factorisations of RHS are 441,1 147,3 49,9 63,7 to give two simultaneous equations for n and k. Solving these gives
n=25 k=55
n=27/4 k=18
n=1 k=5
n=7/4 k=4
So just 2 solutions for integers n and k: n=1 and 25. Two series are 25 = 5^2 (trivial) and 25 + 33 + … + 217 = 3025 = 55^2
Cinelli
Sum of the AP is n*(4n+21) so we are looking for positive integer k for which
4n^2+21*n = k^2
Multiply through by 16 to avoid fractions: 64*n^2 + 16*21*n  16*k^2 = 0
Complete square: (8n+21)^2 – 441 – 16*k^2=0
Factorise difference of two squares: ((8n+21)+4k) * ((8n+21)4k) = 21^2
Possible factorisations of RHS are 441,1 147,3 49,9 63,7 to give two simultaneous equations for n and k. Solving these gives
n=25 k=55
n=27/4 k=18
n=1 k=5
n=7/4 k=4
So just 2 solutions for integers n and k: n=1 and 25. Two series are 25 = 5^2 (trivial) and 25 + 33 + … + 217 = 3025 = 55^2
Cinelli

 Lemon Slice
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Re: Square
OK, now we have had 3 solutions, what about these variations on Cinelli's series?
33 + 41 + 48 + ... + (8n + 17)  first term dropped
17 + 25 + 33 + ... + (8n + 17)  new first term inserted
26 + 34 + 42 + ... + (8n + 18)  first term increased by 1
modellingman
33 + 41 + 48 + ... + (8n + 17)  first term dropped
17 + 25 + 33 + ... + (8n + 17)  new first term inserted
26 + 34 + 42 + ... + (8n + 18)  first term increased by 1
modellingman

 Lemon Quarter
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Re: Square
modellingman wrote:OK, now we have had 3 solutions, what about these variations on Cinelli's series?
33 + 41 + 48 + ... + (8n + 17)  first term dropped
17 + 25 + 33 + ... + (8n + 17)  new first term inserted
26 + 34 + 42 + ... + (8n + 18)  first term increased by 1
modellingman
Those yield to exactly the same approach I already used. The solutions are respectively {50}, {8} and {}.

 Lemon Slice
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Re: Square
UncleEbenezer wrote:modellingman wrote:OK, now we have had 3 solutions, what about these variations on Cinelli's series?
33 + 41 + 48 + ... + (8n + 17)  first term dropped
17 + 25 + 33 + ... + (8n + 17)  new first term inserted
26 + 34 + 42 + ... + (8n + 18)  first term increased by 1
modellingman
Those yield to exactly the same approach I already used. The solutions are respectively {50}, {8} and {}.
Indeed. And you spotted the inconsistency I introduced with n. I should, of course, have written
33 + 41 + 48 + ... + (8n + 25)
17 + 25 + 33 + ... + (8n + 9)
which adjusts your results to {49}, {9} and {}.
modellingman
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