# Particular Diophantine equation [closed]

There are four natural numbers $$a$$, $$b$$, $$c$$ and $$d$$ such that $$a. They satisfy the following equation:

$$a^2$$+$$b^2$$+$$c^2$$+$$d^2$$= $$abcd$$

What is the smallest possible value of $$d$$? This is a math puzzle, so please solve it without computers.

• The answer seems to be $262$, but I really don't see how this can be solved with no-computers. Also it doesn't look like a suitable question on this site. – WhatsUp Oct 25 '20 at 1:49
• A hint: if $(a,b,c,d)$ is a solution, then $(a,b,c,abc-d)$ also. – WhatsUp Oct 25 '20 at 2:00
• @WhatsUp Why do you think this isn't a suitable question for this site? It seems like a fine and valid mathematical puzzle to me. – Rand al'Thor Oct 25 '20 at 22:08
• @WhatsUp Other sites' scope should never be a consideration when deciding whether or not a question is suitable for this site; many SE sites have scope overlaps, including Maths and Puzzling. Not sure if you're familiar with Puzzling's policy on maths problems, but being common in contest maths problems doesn't make something not a puzzle either; quite the contrary, in my opinon, contest maths questions are quite likely to have the "aha" requisite to be on-topic here. – Rand al'Thor Oct 25 '20 at 22:40
• @Randal'Thor Indeed, I have read that meta post and the most voted answer ends with "there's lots of grey area in between." I understand this as depending on personal taste, and this one doesn't look like a puzzle to me... Nevertheless I didn't vote for close. – WhatsUp Oct 25 '20 at 22:46

## 1 Answer

First, let $$2^n$$ be the greatest power of $$2$$ shared among $$a,b,c,d$$ and write $$a=2^n a'$$ etc. Then $$a'^2+b'^2+c'^2+d'^2 = 4^n a'b'c'd'$$. As at least one of the $$a',b',c',d'$$ is odd, the l.h.s. cannot be a multiple of $$8$$ and $$n$$ must be $$1$$ and $$a',b',c',d'$$ all odd.

If we solve for $$d'$$ we get the Pythagorean quintuple $$a'^2 + b'^2 + c'^2 + (d'-2a'b'c')^2 = 4a'^2b'^2c'^2$$. In particular, $$a'^2 + b'^2 + c'^2$$ is of the form $$(2a'b'c')^2 - (2a'b'c'-k)^2 = 4ka'b'c'-k^2$$.

Fortunately, a small solution exists in $$k=a'=1,b'=3,c'=11$$ and this yields $$a=2,b=6,c=22,d=262$$.

Note that formally, $$k$$ and $$d'$$ satisfy the same equation except for the constraint $$d'>a',b',c'$$, so our treatment so far may serve as a derivation of @WhatsUp's transform $$(a,b,c,d) \mapsto (a,b,c,abc-d)$$ which takes solutions (ignoring the $$a constraint) to solutions.

So another way of looking at this is: Partially ignore that $$a,b,c,d$$ must be distinct. Guess solution $$a',b',c',d' = 1,1,3,11$$ or $$a,b,c,d = 2,2,6,22$$ and apply WhatsUp's transformation to make the numbers distinct.

We can take this two steps further: Start from $$a,b,c,d = 2,2,2,2$$ and apply WhatsUp transform three times: $$(2,2,2,2)\rightarrow(2,2,2,6)\rightarrow(2,2,6,22)\rightarrow(2,6,22,262)$$