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For which $k$ do there exist $k$ pairwise distinct primes $p_{1}, p_{2}, ..., p_{k}$ such that $(p_{1})^2 + .... + (p_{k})^2 = 2010$

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`2^{34}`	$2^{34}$
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Comments

I didn"t get (pairwise distinct primes ?)

A Former Brilliant Member - 5 years, 6 months ago

1²+2²+...+29²=2398 So k<11 Also 43²=1849,47²=2209 So Pk<43 Pk=37 also can't be because 2010-37²=641 but we check it can"t be sum Pk=31 ,2010-961=1049 1049=29²+13²+5²+3²+2² So one solution is 31,29,13,5,3,2 So for Pk=29 2010-29²=1269 We know that 2²+3²+...+23²=1557 So 1557-1269=288 But it can't be the sum So there is no more solutions becaus1557<2010 Only solution is 2²+3²+5²+13²+29²+31²=2010,so k=6

Nikola Djuric - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$