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# Help!

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

Note by Dev Sharma
1 year, 3 months ago

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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 · 1 year, 3 months ago