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How to calculate the multiplying of the three numbers $(n_1, n_2, n_3)$ where the summation of them equals 784 if you know the followings: $n_1, n_2, n_3$ are natural numbers $n_1 < n_2 < n_3$, $n_1^2 + n_2^2 = n_3^2$

N. F. Taussig
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omar.sayed
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2 Answers2

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For our purposes we can relax the constraint $n_1<n_2$ and assume $n_1=2dab,\,n_2=d(a^2-b^2),\,n_3=d(a^2+b^2)$ for naturals $a,\,b,\,d$ with $a>b,\,(a,\,b)=1$, so $da(a+b)=392=2^3 7^2$. We need $a+b<2a$ and $(a,\,a+b)=1$, so $a+b=2^3=8,\,a=7,\,b=1,\,d=7$ and $n_1 n_2 n_3 = 2\times 7^4\times (7^4-1^4)=11524800$.

J.G.
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  • If the question had $56$ instead of $784$, would you have got $d=\frac12$? – Henry Aug 09 '17 at 17:25
  • @Henry No: in that case $d=1,,a=4,,b=3$. – J.G. Aug 09 '17 at 17:46
  • OK, though that might not satisfy your $a+b<2a$. Relaxing that requirement could give $d=14, a=4, b=3$ as a solution in the $784$ case – Henry Aug 09 '17 at 17:56
  • @Henry Firstly, $4+3< 2\times 4$. Secondly, your proposed factorisation gives the same $n_1 n_2 n_3$, but thanks for mentioning it. – J.G. Aug 09 '17 at 18:01
  • My mistake on the first point. There is essentially one solution to the original question, but I had found it peculiar that a similar triangle was not giving the same $a$ and $b$ – Henry Aug 09 '17 at 18:06
  • @Henry That's due to a subtlety in how Pythagorean triples work when you double or halve the numbers, viz. $a'=a+b,,b'=a-b$ or the inverse transformation. This makes one parametrisation universal even though it seems at first as if parity concerns establish two. – J.G. Aug 09 '17 at 18:19
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Hint: $7,24,25$ is a primitive Pythagorean triple. Note that $7+24+25 = 56$ divides 784.

Math Lover
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