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    Prove algebraically that the sum of two squares of two consecutive integers is always an odd number.
    How can you prove it algebraically when it refers to integers?
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    (Original post by sanibonani)
    Prove algebraically that the sum of two squares of two consecutive integers is always an odd number.
    How can you prove it algebraically when it refers to integers?
    For an integer n, the next one along is n+1. So, now translate the part "sum of two squares of two consecutive integers" into mathematical form and proceed from there to show that this result is 1 more/less than a multiple of 2.
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    Going into more detail from RDKGame's reply, take the integer as n and the next as n+1 (as RDKGames says). Then square each so you have n^2 and (n+1)^2. Then add those two terms so you get n^2 + (n+1)^2 which = n^2 + n^2 + n + n + 1 = 2n^2 + 2n + 1 = 2(n^2 + n) + 1
    Multiples of 2 are always even so 2(n^2 + n) must be even
    An even number plus an odd number must be odd so 2(n^2 + n) + 1 must be odd.
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    (Original post by ellie154352)
    Going into more detail from RDKGame's reply, take the integer as n and the next as n+1 (as RDKGames says). Then square each so you have n^2 and (n+1)^2. Then add those two terms so you get n^2 + (n+1)^2 which = n^2 + n^2 + n + n + 1 = 2n^2 + 2n + 1 = 2(n^2 + n) + 1
    Multiples of 2 are always even so 2(n^2 + n) must be even
    An even number plus an odd number must be odd so 2(n^2 + n) + 1 must be odd.
    Correct but I believe posting full solutions on the forum is discouraged, rather you should give hints and tips to encourage working.
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    (Original post by mundosinfin)
    Correct but I believe posting full solutions on the forum is discouraged, rather you should give hints and tips to encourage working.
    Will keep that in mind (wasn't really aware of it). Thanks
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    Thanks for the replies but I've figured it out now
 
 
 
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