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DRAFT
x|y| means that x divides y.
1. Clearly, c^2={0,1} mod4
2. Also, a^4+b^4+1={1,2,3} mod4; accordingly, 2|a|,2|b|
3. Let a=2r; let b=2s: now, (2r)^4+(2s)^4+1={0,3,7} mod10;5|r|,5|s|
4. Taking out the factors 2 and 5, we obtain for some x,y: 2^4*5^4(x^4+y^4) = c^2-1;
This restriction forces c to take one of four forms: c=5000n+{1,1249,3751,4999}
[In the remainder of the proof we will successively substitute on either side for each of these forms, with the idea of reaching a contradiction for each case of each form.]
5. From 4. above we have: a^4+b^4=10000(x^4+y^4);
(i.) Let c = 5000n+1; then c^2=25000000n^2+10000n+1=10000n(2500n+1)+1
Simplifying by deducting 1 from this and cancelling: x^4+y^4=n(2500n+1)
(ii.)Let c = 5000n+1249; then c^2=25000000n^2+12490000n+1560001=10000(2500n^2+1249n+156)+1
Simplifying by deducting 1 from this and cancelling: x^4+y^4=2500n^2+1249n+156=(4n+1)(625n+156)
(iii.)Let c = 5000n+3751; then c^2=25000000n^2+37510000n+14070001=10000(2500n^2+3751n+1407)+1
Simplifying by deducting 1 from this and cancelling:
x^4+y^4=2500n^2+3751n+1407=(4n+3)(625n+469)
(iv.)Let c = 5000n+4999; then c^2=25000000n^2+49990000n+24990001=10000(2500n^2+4999n+2499)+1
Simplifying by deducting 1 from this and cancelling:
x^4+y^4=(2500n^2+4999n+2499)=(n+1)(2500n+2499)
[Now we are going to consider the forms that c can take in terms of a final variable, p]
6. It is also clear that c^2=1 mod16. It follows that c itself must be of the form {c=8p+7,8p-7)}
(i.) Let c = 8p-7: expanding: a^4+b^4+1=64p^2-112p+49; deducting 1 from both sides: a^4+b^4=16(p-1)(4p-3). We now substitute for each of the four cases discussed above:
case(a) 16(p-1)(4p-3)=10000n(2500n+1): simplifying; p=625n+1
case(b) 16(p-1)(4p-3)=10000(2500n^2+1249n+156): simplifying p=625n+157
case(c) 16(p-1)(4p-3)=10000(2500n^2+3751n+1407): simplifying; 4p=2500n+1879:
leading to the contradiction that p is a fraction, so this case need not be further considered
case(d) 16(p-1)(4p-3)=10000(2500n^2+4999n+2499);4p=2500 n+2503:
leading to the contradiction that p is a fraction, so this case need not be further considered
(ii.)Let c=8p+7: a^4+b^4+1=64p^2+112p+49;a^4+b^4=16(p+1)(4p+3)
case(a) 16(p+1)(4p+3)=10000n(2500n+1); 4p=2500n-3:
leading to the contradiction that p is a fraction, so this case need not be further considered
case(b) 16(p+1)(4p+3)=10000(2500n^2+1249n+156); 4p=2500n+621:
leading to the contradiction that p is a fraction, so this case need not be further considered
case(c) 16(p+1)(4p+3)=10000(2500n^2+3751n+1407);p=625n+468
case(d) 16(p+1)(4p+3)=10000(2500n^2+4999n+2499);p=625n+624
[There remain four cases not yet eliminated by contradiction:- 6.(i.) cases (a) and (b) plus 6.(ii.) cases (c) and (d). We are going to substitute the values we obtained in 6. above for p in terms of n back into the equation for c^2 in terms of p. For good measure, we will also cross-check against the corresponding results in 5. above]
7(i.)c^2=16(p-1)(4p-3)+1,
(a)Let p=625n+1: substituting;
c^2=16(625n)(4*625n-2)+1; expanding: c^2=10000n(2500n-2)+1 =[from 6.(i.)(a)]10000n(2500n+1)+1; which is true iff n=0
checking:10000n(2500n-2)+1=10000(x^4+y^4)+1=[from 5.(i.)]10000n(2500n+1)+1;
and again, n=0
(b)Let p=625n+157: substituting;
c^2=16(625n+157-1)(4*625n+157-3)+1: expanding; c^2=25000000n^2+7780000n+384385 =[from 6.(i.)(b)]25000000n^2+12490000n+1560001; which is true iff n is a (negative) fraction, -156/625
checking:25000000n^2+7780000n+384385=[from 5.(ii.)]10000(4n+1)(625n+156)+1
and again, n is a (negative) fraction
7(ii.) c^2=16(p+1)(4p+3)+1
(a)Let p=625n+468: substituting;
c^2=16(625n+468+1)(4*625n+468+3)+1: expanding; c^2=25000000n^2+23470000n+3534385
=[from 6.(ii.)(c)]25000000n^2+37510000n+14070001; which is true iff n is a (negative) fraction, -469/625
checking:25000000n^2+23470000n+3534385=[from 5.(iii.)]10000(4n+3)(625n+469)+1
and again, n is a (negative) fraction
(b)Let p=625n+624: substituting;
c^2=16(625n+624+1)(4*625n+624+3)+1; expanding: c^2=25000000n^2+31270000n+6270001
=[from 6.(ii.)(d)]25000000n^2+49990000n+24990001
which is true iff n=-1
checking 25000000n^2+31270000n+6270001=[from 5.(iv.)]10000(n+1)(2500n+2499)+1
confirms that n=-1
It follows that there is no solution in the positive integers for the equation a^4+b^4+1 = c^2 |
