Find the distinct positive integers $a, b, c,d$, such that the following conditions hold: (1) exactly three of the four numbers are prime numbers; (2) $a^2 + b^2 + c^2 + d^2 = 2018.$
Problem
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Tags: number theory, Sum of Squares
lksb
05.09.2024 05:07
No solutions?
Let $d$ be the composite number
$a^2+b^2+c^2=(45+d)(45-d)-7$
By $\pmod{6}$ we have $$3=9-d^2-7=2-d^2 \to d^2=5$$Which is a contradiction unless $2=a\leq b \leq c$
Therefore, $a=2$ and $d^2\equiv 1\pmod{6}$
It follows $$b^2+c^2+(6k\pm1)^2=2014\to b^2+c^2+(6k)^2=2013\pm 12k=b^2+c^2+t^2=2013\pm2t$$By $\pmod 2$, we have that $b=2$
Then, we have that $$c^2+d^2=2010=2\cdot 3\cdot 5\cdot 67$$A contradiction by Thue's Lemma
lksb
06.09.2024 02:45
I claim that the solution is $(a,b,c,d)=(2,3,41,18)$ Proof: Let $d$ be the composite number By mod 6 we have $d^2\pmod6=5$, a contradiction, therefore, $\fbox{a=2}$ By mod 3 we have $d^2\pmod3=5$, a contradiction, therefore, $\fbox{b=3}$ It follows: $$c^2+d^2=2005$$$$c=43\implies d^2=2005-1849=156$$$$c=41 \implies d=18$$Therefore, $$\fbox{(a,b,c,d)=(2,3,41,18)}$$