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Four different positive integers $a, b, c$, and $d$ are such that $a^2 + b^2 = c^2 + d^2$ What is the smallest possible value of $abcd$?

I just need a few hints, nothing else. How should I begin? Number theory?

Andrés E. Caicedo
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Robart
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  • The formula there. http://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 – individ Jan 24 '15 at 17:19
  • We used that problem in round 2 of this year's Abel competition. If you can read Norwegian, the solution is on the web site. (It is there even you don't read Norwegian, of course. You may be able to get something out of the solution even so.) – Harald Hanche-Olsen Jan 24 '15 at 17:41
  • @HaraldHanche-Olsen, interestingly, i lived in Norway for 4 years, fluent in Norwegian, but I lost the hang of it, I cant understand the solution there. I got this problem from there too! – Robart Jan 24 '15 at 17:43
  • @HaraldHanche-Olsen, Its a great contest. When I lived in Norway, I couldnt take it because it wasnt available. There are still some people from my school, Oslo International School, who are doing well in it. Im 16 now, I was 13 back then, and my school didnt offer it =) – Robart Jan 24 '15 at 17:46

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The number of representations of a positive integer as a sum of two squares depends on the number of prime divisors of the form $4k+1$ (see Cox, Primes of the form $a^2+k b^2$). If we take the first two primes of such a form, $5$ and $13$, we have that $5\cdot 13$ can be represented as: $$ 65 = 1^2+8^2 = 4^2+7^2 $$ so we have a solution with $abcd = 224$. You can complete the proof by exhaustive search (it is quite easy to check that the first $64$ positive integers do not have a double representation in terms of positive integers, and we just have to check till $n=225$ or so to find the minimum $abcd$), or proving that the number of representations of $n$ as a sum of two squares is given by the number of divisors of $n$ of the form $4k+1$ minus the number of divisors of the form $4k+3$.

Jack D'Aurizio
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  • Nice Method, where can I get introductor information on this? – Robart Jan 24 '15 at 17:33
  • @Robart: the book of D.H.Cox is a masterpiece, I suggest you to have a look at it. – Jack D'Aurizio Jan 24 '15 at 17:35
  • But why should the solution with the smallest $n$ also be the solution with the smallest $abcd$? – vuur Jan 24 '15 at 17:46
  • @vuur: Ok, it is not right, but we just need to check till $n=15^2$ or so, it is not much expensive. The next solution is $85=5\cdot 17 = 2^2+9^2 = 6^2+7^2$ with a much bigger $abcd$ yet. – Jack D'Aurizio Jan 24 '15 at 17:49
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    Yes, you are right, that's a simple and inexpensive bound. I was just being pedantic :p – vuur Jan 24 '15 at 17:59
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    Actually, one only need to check up to $n = 114$.

    WOLOG, assume $a < b, a < c, c < d$. If $a^2 + b^2 = c^2 + d^2 = n$, then

    $$\begin{array}{rll} a \ge 1 &\implies a^2 b^2 \ge 1(n-1) &\implies ab \ge \sqrt{n-1}\ c \ge 2 &\implies c^2 d^2 \ge 4(n-4) &\implies cd \ge 2\sqrt{n-4} \end{array}$$

    If $abcd < 224$, then $2\sqrt{(n-1)(n-4)} < 224 \implies n \le 114$. Below $115$, there are only 3 numbers: $50$,$65$ and $80$ which have multiple representation as sum of two squares.

     – achille hui Jan 24 '15 at 18:28