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The series of Find a factor puzzle is started by Culver Kwan, and asks the solver to identify a factor of a certain large number within a certain range using some mathematical identities. This should be tagged with , and .


Find a factor of $104060405$ within $10100$ and $11000$, using various mathematical identities. You should not use a computer.

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  • $\begingroup$ Does seeing a factorization count as using an identity? This factors easily if you rot13(oernx vg va gur zvqqyr), so to speak. $\endgroup$ – msh210 May 25 at 3:45
  • $\begingroup$ @msh210 I do not understand what are you saying, but the identity is a Wikipedia page. $\endgroup$ – Culver Kwan May 25 at 4:10
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To be honest, like msh210 said in the comments, I just saw the factorisation:

$104060405 = 104050000 + 10405 = 10405*(10000+1) = 10405*10001$

You could find it by recognising

most of the fourth row of Pascal's triangle $1 4 6 4 1$, and using
$(x+1)^4=x^4+4x^3+6x^2+4x+1$
with $x=100$ to write the number as
$104060405=101^4+2^2=10201^2+2^2$

And then applying the identity

for the product of sums of two squares:
$10201^2+2^2 = (102*100+1*1)^2+(102*1-100*1)^2 = (102^2+1^2)(100^2+1^2)$

but that seems a bit convoluted to me.

Edit:

The intended trick was to use:

Sophie Germain's identity which is
$x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$
In this case we have $x=101$ and $y=1$, so the factors are
$101^2\pm 202 + 2 = 10203\pm 202$

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  • $\begingroup$ Do you know the Sophie Germain's identity? It is in a small section of a Wikipedia page, but you can see the explanation in other sites. I used this identity to create this problem. Still, your answer is good and a check mark will be rewarded. I will try not to make a question which can be done with very simple factorisation next time. $\endgroup$ – Culver Kwan May 25 at 13:20
  • $\begingroup$ @CulverKwan I was not really aware of that identity. I have no doubt seen it before, but it is not one that I have ever needed so it is not in my arsenal as it were. $\endgroup$ – Jaap Scherphuis May 25 at 14:10
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The number trivially splits into

$5 \times 20812081 = 5 \times 10001 \times 2081 = 10001 \times 10405$

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