Wolfram Alpha gives me a=31, b=41, and c=49, but how is this done by hand? If it helps: try visualizing it on a number line. Thanks!
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$b^2 - a^2 = (b - a)(b + a)$. If $b-a = d$ and $b+a = e$, $a = (e-d)/2$ and $b = (d+e)/2$. These will be integers if $d$ and $e$ are integers that are
congruent mod $2$, i.e. both odd or both even; $a$ is half their difference and $b$ is their average.
Now in this case you want to factor $720$ in two ways, say $de$ corresponding to $a$ and $b$ as above, and $fg$ corresponding to $b$ and $c$. These have $b$ in common, i.e. the difference between $f$ and $g$ is the sum of $d$ and $e$. The ways of factoring $720$ into two numbers congruent mod $2$, and the sums and (positive) differences, are as follows:
$$ \matrix{d & e & \text{sum} & \text{difference}\cr
2 & 360 & 362 & 358\cr
4 & 180 & 184 & 176\cr
6 & 120 & 126 & 114\cr
8 & 90 & 98 & 82\cr
10 & 72 & 82 & 62\cr
12 & 60 & 72 & 48\cr
18 & 40 & 58 & 22\cr
20 & 36 & 56 & 16\cr
24 & 30 & 54 & 6\cr
}$$
The only number that appears in both the sum and difference columns is $82 = 10 + 72 = 90 - 8$. Thus the only solution is $a = (72 - 10)/2 = 31$,
$b = (72+10)/2 = 41$, $c = (90 + 8)/2 = 49$.
hmakholm left over Monica
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Robert Israel
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Just curious. How come you only checked those factors? – Jacob Wheeler Apr 03 '15 at 22:32
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I checked all factors $d \le \sqrt{720}$ such that $720/d \equiv d \mod 2$. – Robert Israel Apr 03 '15 at 22:48
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But why choose only evens? – Jacob Wheeler Apr 03 '15 at 23:03
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1As I stated, $a = (c-d)/2$, so for this to be an integer $c$ and $d$ must be congruent mod $2$. Since $720$ is even, they can't both be odd, so they must both be even. – Robert Israel Apr 04 '15 at 00:23
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Using the difference of squares formula, we get that:
$$(b-a)(b+a) = 720$$
and that:
$$(c-b)(c+b) = 720$$
Now, note the prime factorization of $720$: $2^4*3^2*5$
Therefore, each pair of factors of $720$ correspond to, for example, in the case of $b^2 - a^2$, $(b-a)$ and $(b+a)$.
Therefore, if you test out the cases you will arrive at your answer: $a = 31; b=41; c = 49$
Varun Iyer
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So you must go through all of the factor pairs until you get 720? – Jacob Wheeler Apr 03 '15 at 22:06
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@JacobWheeler I wouldn't suggest doing that. Note that you have $2$ equations where the variable $b$ is common in both of them. That should help to eliminate some of the cases you have to consider – Varun Iyer Apr 03 '15 at 22:06
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Hint: Substitute $b^2= 720+a^2$ into the second equation to get $$c^2-a^2 = (c+a)(c-a) = 2 \cdot 720.$$
Sandeep Silwal
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Each perfect square $n^2$ is the sum of the first $n$ odd numbers, so the difference between two squares is always the sum of a contiguous sequence of odd numbers.
The sum of such an arithmetic progression is the number of terms times the mean of the first and last terms. If there's an odd number of terms, then the mean term is just the middle term, which is also odd, and so the sum of the progression is odd. Thus, that case cannot give us a 720 as a difference between squares.
On the other hand, if there's an even number of terms, then the mean term is between the two middle term, and is therefore an even number. So in order to find two squares with a difference of 720, we must find two even numbers that multiply to 720. Each such multiplication directly gives us the mean term and the number of terms, so we can figure out where the sequence starts and ends.
There are 9 ways to write 720 as a product of even numbers. Compute the odd-number-sequence is for each of them, and look for one that begins just after another one ended.
hmakholm left over Monica
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