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    1) If x and y are prime numbers, solve the equation x^y - y^x = x.y^2 - 19.

    The main tool is Fermat's little theorem: if p is prime then
    a^p=a mod p. So (i)x divides y-19, and

    (ii) y divides x+19, and in particular y<=x+19

    Suppose y>19. By (i),x<=y-19 so by (ii) x=y-19. But
    this is impossible (one of x,y would have to be even, hence x=2,y=21).

    If y=19, then by (ii) x=19, which doesn't work.

    So y<19 and by (i) x divides 19-y. This gives the
    following possibilities:

    y=2, x=17

    y=3, x=2

    y=5,x=2 or 7

    y=7, x=2 or 3


    y=13,x=2 or 3


    Taking (ii) into account reduces the possibilities to three:



    One checks by hand that the first doesn't work, but the other two do.
    So there are exactly two solutions

    2) find n so (1+4+9+16+....+n^2)/n is a square number

    We want to make (n+1)(2n+1)/6 a square. First, this is not usually even an integer. As 2n+1 is odd, n+1 must be even, say n=2k+1. This gives (k+1)(4k+3)/3. So 3 must divide k or k+1.

    Suppose k+1=3m. Then m(12m-1) is a square. As a prime p cannot divide both m and 12m-1, both m and 12m-1 must be squares, say m=r2, 12m-1=s2. Then 12r2=s2+1. Then s must be odd, say s=2p+1 and 6r2=2p2+2p+1. But the right-hand side is never divisible by 3 (consider mod 3). So this case is impossible.

    Hence k=3m and (3m+1)(4m+1) must be a square. If a prime p divides 3m+1 and 4m+1, it divides m, so would not divide 3m+1. Hence 3m+1 and 4m+1 must both be squares, say 3m+1=p2, 4m+1=q2. Let q=p+r. Then m=2pr+r2 and p2-6pr-3r2-1=0. This gives p=
    3r+√12r2+1 (the other solution is negative) and n=42r2+12r√(12r2+1)+1. As n increases with r, the smallest solution for n corresponds to the smallest value of r that makes this expression for n an integer. This value is r=2, giving n=337.

    wow, Fermat and Galois all on one forum

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